Exercise 2.1
अतिलघु उत्तरीय प्रश्न(Very Short Answer Type Questions)
TYPE-I परिमेय एवं अपरिमेय संख्याओ के गुणों पर आधारित प्रश्न
Question 1
बताइए कि नीचे दी गई संखयाओ में कौन-कौन परिमेय हैं और कौन-कौन अपरिमेय है:(i) 3+√5
Sol : अपरिमेय
(ii) 7√5
Sol : अपरिमेय
(iii) \dfrac{7}{\sqrt{5}}
Sol : अपरिमेय
(iv) 5+√5-√5
Sol : परिमेय
(v) √2+21
Sol : अपरिमेय
(vi) π-2
Sol : अपरिमेय
(vii) √2 + √3
Sol : अपरिमेय
Question 2
निम्नलिखित में से किसका हल परिमेय संख्याओ को निरुपित करते है ?(i) x2=5
Sol :
x=±√5
(ii) x^2=\dfrac{16}{9}
Sol :
x^2=\pm \dfrac{16}{9}
अपरिमेय
(ii) x^2=\dfrac{16}{9}
Sol :
x^2=\pm \dfrac{16}{9}
x=\pm \dfrac{4}{3}
(iii) (x+√2)(x-√3)=0
Sol :
x=-√2 , x=-√3
(iv) 3x2=9
Sol :
x^2=\dfrac{9}{3}
x2=3
x=±√3
(v) \sqrt{3}x=\dfrac{3}{4}
Sol :
x=\dfrac{3}{4\sqrt{3}}
x=\dfrac{3}{4\sqrt{3}}\times \dfrac{\sqrt3}{\sqrt3}
x=\dfrac{\sqrt{3}}{4}
अपरिमेय
(vi) x^2=\dfrac{25}{49}
Sol :
⇒x=\pm \sqrt{\dfrac{25}{49}}
⇒\pm \dfrac{5}{7}
(vii) (x+1)2=6
Sol :
⇒x+1=±√6
⇒x=±(√6)-1
⇒x=+√6-1 , -√6-1
अपरिमेय
(viii) (x+√5)(x-√3)=0
Sol :
परिमेय
Sol :
x=-√2 , x=-√3
अपरिमेय
Sol :
x^2=\dfrac{9}{3}
x2=3
x=±√3
अपरिमेय
Sol :
x=\dfrac{3}{4\sqrt{3}}
x=\dfrac{3}{4\sqrt{3}}\times \dfrac{\sqrt3}{\sqrt3}
x=\dfrac{\sqrt{3}}{4}
अपरिमेय
Sol :
⇒x=\pm \sqrt{\dfrac{25}{49}}
⇒\pm \dfrac{5}{7}
परिमेय
Sol :
⇒x+1=±√6
⇒x=±(√6)-1
⇒x=+√6-1 , -√6-1
अपरिमेय
Sol :
x=-√5 , +√5
(i) x2=5
Sol :
⇒x=±√5
(ii) x^2=\dfrac{16}{9}
Sol :
⇒x=\sqrt{\dfrac{16}{9}}
⇒x=\pm \dfrac{4}{3}
(iii) (x-1)^2=\dfrac{49}{16}
Sol :
⇒x-1=\sqrt{\dfrac{49}{16}}
⇒x-1=\pm \dfrac{7}{4}
⇒x-1= +\dfrac{7}{4} and x-1=-\dfrac{7}{4}
⇒x=\dfrac{7}{4}+1 and x=-\dfrac{7}{4}+1
⇒x=\dfrac{7+4}{4} and x=\dfrac{-7+4}{4}
⇒x=\dfrac{11}{4} and x=\dfrac{-3}{4}
(iv) (x+1)(x-1)=0
Sol :
⇒x2-1=0
⇒x=±√1
(v) x^2=\dfrac{19}{29}
Sol :
⇒x=\sqrt{\dfrac{19}{29}}
(iii) अंतर परिमेय संख्या है ।
2√3×√3=6
2√3,√3
(vi) गुणनफल अपरिमेय संख्या है ।
Sol :
2√3×2√2=4√6
2√3,2√2
(vii) भागफल परिमेय संख्या है ।
Sol :
\frac{2\sqrt{3}}{\sqrt3}
2√3,√3
(viii) भागफल अपरिमेय संख्या है ।
Sol :
\frac{4\sqrt{6}}{\sqrt{5}}
4√6,√5
Sol :
⇒0 , 2√3
परिमेय , अपरिमेय
⇒0×2√3=0 परिमेय
(i) (5+√5)(5-√5)
Sol :
⇒(5)2-(√5)2
⇒25-5
⇒20
(ii) (5+√7)(2+√5)
Sol :
⇒5(2+√5)+√7(2+√5)
⇒10+5√5+2√7+√35
निम्नलिखित मे से प्रत्येक का सरलतम परिमेयकारी गुणक लिखे:
(i) 5√2
Sol :
√2
(ii) 2√2
Sol :
√2
(iii) √7
Sol :
√7
(iv) √15
Sol :
√15
यदि a,b,c परिमेय संखयाएँ हो तो (i) \sqrt[5]{a^2b^3c^4} तथा (ii) \sqrt[9]{a^2b^4c^8} का परिमेयकारी गुणक ज्ञात करे ।
Sol :
(i) \sqrt[5]{a^3b^2c}
(ii) \sqrt[9]{a^7b^5c}
निम्नलिखित मे प्रत्येक के हर को परिमेय बनाकर लिखे:
(i) \frac{1}{\sqrt2}
Sol :
हर का परिमेयकरण करने पर,
\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}
(ii) \frac{1}{\sqrt{12}}
Sol :
\frac{1}{\sqrt{12}}=\frac{1}{\sqrt{2\times 2 \times 3}}
=\frac{1}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
=\frac{\sqrt{3}}{2\times 3}=\frac{\sqrt{3}}{6}
(iii) \frac{2\sqrt{7}}{\sqrt{11}}
Sol :
\frac{2\sqrt{7}}{\sqrt{11}}\times \frac{\sqrt{11}}{\sqrt{11}}
=\frac{2\sqrt{77}}{11}
(iv) \frac{2}{\sqrt{17}}
Sol :
रिक्त स्थानो मे हर को परिमेय बनाकर लिखो:
(i) \frac{1}{\sqrt{2}+1}=\dots
Sol :
हर का परिमेयकरण करने पर,
\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}
=\frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}
=\frac{\sqrt{2}-1}{2-1}=\frac{\sqrt{2}-1}{1}
=\sqrt{2}-1
(ii) \frac{1}{2-\sqrt{3}}=\dots
Sol :
(iii) \frac{3}{\sqrt{5}+\sqrt{3}}=\dots
Sol :
(iv) \frac{7}{\sqrt{5}-\sqrt{3}}=\dots
Sol :
निम्नलिखित करणियो को सरलतम रुप मे लिखे:
(i) √48
Sol :
=√2×2×2×2×3
=2×2√3
=4√3
(ii) √175
Sol :
(iii) ∛72
Sol :
=∛2×2×2×3×3
=2×∛9 या
=∛2×2×2×3×3×3×\frac{1}{3}
=6\sqrt[3]{\frac{1}{3}}
(iv) √125
Sol :
(v) ∛54
Sol :
(vi) ∛144
Sol :
(vii) \sqrt[5]{320}
Sol :
(viii) \sqrt{\frac{125}{63}}
Sol :
(ix) \sqrt{\frac{112}{45}}
Sol :
=\sqrt{\frac{2\times 2 \times2 \times 2\times 7 }{3\times 3 \times 5}}
=\frac{4}{3}\sqrt{\frac{7}{5}}
निम्नलिखित मे से प्रत्येक के हर का परिमेयकरण करे :
(i) \frac{1}{2+\sqrt{3}}
Sol :
(ii) \frac{1}{7+3\sqrt{2}}
Sol :
=\frac{1}{7+3\sqrt{2}}\times \frac{7-3\sqrt{2}}{7-3\sqrt{2}}
=\frac{7-3\sqrt{2}}{(7)^2-(3\sqrt{2})^2}
=\frac{7-3\sqrt{2}}{49-18}
=\frac{7-3\sqrt{2}}{31}
(iii) \frac{5}{\sqrt{3}-\sqrt{5}}
(iv) \frac{6}{3\sqrt{2}-2\sqrt{3}}
Sol :
=\frac{6}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^2-(2\sqrt{3})^2}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{18-12}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{6}
=3√2+2√3
(v) \frac{4}{\sqrt{5}+\sqrt{3}}
Sol :
(vi) \frac{5+\sqrt{6}}{5-\sqrt{6}}
Sol :
=\frac{5+\sqrt{6}}{5-\sqrt{6}}\times \frac{5+\sqrt{6}}{5+\sqrt{6}}
=\frac{(5+\sqrt{6})(5+\sqrt{6}}{(5)^2-(\sqrt{6}^2}
=\frac{25+5\sqrt{6}+5\sqrt{6}+6}{25-6}
=\frac{31+10\sqrt{6}}{19}
(vii) \frac{3+\sqrt{2}}{3-\sqrt{2}}
(viii) \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}
Sol :
=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times \frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}
=\frac{6\sqrt{30}+4\times 6-3\times 5-2\sqrt{30}}{45-24}
=\frac{4\sqrt{30}+19}{21}
(ix) \frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}
Sol :
निम्नलिखित को सरल करे:
(i) \frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}
Sol :
हर का परिमेय करण करने पर,
=\frac{3}{5-\sqrt{3}}\times \frac{5+\sqrt{3}}{5+\sqrt{3}}+\frac{2}{5+\sqrt{3}}\times \frac{5-\sqrt{3}}{5-\sqrt{3}}
=\frac{3(5+\sqrt{3})}{(5)^2-(\sqrt{3})^2}+\frac{2(5-\sqrt{3})}{(5)^2-(\sqrt{3})^2}
=\frac{3(5+\sqrt{3})}{25-3}+\frac{2(5-\sqrt{3})}{25-3}
=\frac{3(5+\sqrt{3})}{22}+\frac{2(5-\sqrt{3})}{22}
=\frac{3(5+\sqrt{3}+2(5-\sqrt{3}))}{22}
=\frac{15+3\sqrt{3}+10-2\sqrt{3}}{22}
=\frac{25+\sqrt{3}}{22}
(ii) \frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}
Sol :
=\frac{(\sqrt{5}-2)(\sqrt{5}-2)-(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}+2)(\sqrt{5}-2)}
=\frac{(5-2\sqrt{5}-2\sqrt{5}+4)-(5+2\sqrt{5}+2\sqrt{5}+4)}{(\sqrt{5})^2-(2)^2}
=\frac{(9-4\sqrt{5})-(9+4\sqrt{5})}{5-4}
=\frac{9-4\sqrt{5}-9-4\sqrt{5}}{1}
=-8√5
(iii) \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}
Sol :
=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}
=\frac{5+\sqrt{15}+\sqrt{15}+3+5-\sqrt{15}-\sqrt{15}+3}{(\sqrt{5})^2-(\sqrt{3})^2}
=\frac{16}{5-3}=\frac{16}{2}=8
(iv) \frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}+\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}
Sol :
=\frac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})+(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}
=\frac{(\sqrt{5}-\sqrt{3}+\sqrt{10}-\sqrt{6}+\sqrt{5}+\sqrt{3}-\sqrt{10}-\sqrt{6})()+()()}{(\sqrt{5})^2-(\sqrt{3})^2}
=\frac{2\sqrt{5}-2\sqrt{6}}{5-3}
=\frac{2(\sqrt{6}-\sqrt{6})}{2}
=√5-√6
निम्नलिखित को सरल करे:
(i) \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\sqrt{35}
Sol :
हर का परिमेयकरण करने पर,
=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\sqrt{35}
=\frac{7-\sqrt{35}-\sqrt{35}+5}{(\sqrt{7})^2-(\sqrt{5})^2}+\sqrt{35}
=\frac{12-2\sqrt{35}}{7-5}+\sqrt{35}
=\frac{2(6-\sqrt{35})}{2}+\sqrt{35}
=6-√35+√35
=6
(ii) \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+2\sqrt{6}
Sol :
हर का परिमेयकरण करने पर,
=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+2\sqrt{6}
=\frac{3-\sqrt{6}-\sqrt{6}+2}{(\sqrt{3})^2-(\sqrt{2})^2}+2\sqrt{6}
=\frac{5-2\sqrt{3}}{3-2}+2\sqrt{6}
=5-2√6+2√6
=5
(i) यदि a=3+√8 तो a^2+\frac{1}{a^2} का मान ज्ञात करे ।
Sol :
a=3+√8
\frac{1}{a}=\frac{1}{3+\sqrt{8}}
=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}
=\frac{3-\sqrt{8}}{3^2-(\sqrt{8})^2}
=\frac{3-\sqrt{8}}{9-8}
\frac{1}{a}=3-\sqrt{3}
a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2.a.\frac{1}{a}
=(3+\sqrt{8}+3-\sqrt{8})^2-2
=(6)2-2
=36-2
=34
(ii) यदि a=\frac{\sqrt{2}+1}{\sqrt{2}-1}, b=\frac{\sqrt{2}-1}{\sqrt{2}+1} , सिद्ध करे कि a2+b2+ab=35
Sol :
a+b=\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}
=\frac{(\sqrt{2}+1)(\sqrt{2}+1)+(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}
=\frac{2+\sqrt{2}+\sqrt{2}+1+2-\sqrt{2}-\sqrt{2}+1}{(\sqrt{2})^2-(1)^2}
=\frac{6}{2-1}=6
L.H.S
a2+b2+ab=a2+b2+ab
=(a+b)2-2ab+ab
=(a+b)2-ab
=6^2-\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}-1}{\sqrt{2}+1}
=36-1
=35
(iii) यदि a=2+√3 तो a^3+\frac{1}{a^3} का मान ज्ञात करे।
Sol :
a=2+√3
\frac{1}{a}=\frac{1}{2+\sqrt{3}}
=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}
=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}
\frac{1}{a}=2-\sqrt{3}
a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2-a.\frac{1}{a}+\frac{1}{a^2}\right)
=\left(a+\frac{1}{a}\right)\left((a+\frac{1}{a})^2\right)-2a.\frac{1}{a}-1
=4\left[(4)^2-3\right]
=4[16-3]
=4×13
=52
(iv) यदि x=\frac{\sqrt{3}+1}{2} तो 4x3+2x2-8x+7 का मान ज्ञात करे ।
Sol :
x^2=\left(\frac{\sqrt{3}+1}{2}\right)^2
=\frac{(\sqrt{3}^2+2.\sqrt{3}.1+1^2)}{4}
=\frac{4+2\sqrt{3}}{4}=\frac{2(2+\sqrt{3})}{4}
x^2=\frac{2+\sqrt{3}}{2}
x^3=\left(\frac{\sqrt{3}+1}{2}\right)^2
=\frac{(\sqrt{3})^3+3(\sqrt{3})^2.1+3\sqrt{3}.1^2+1^3}{8}
=\frac{3\sqrt{3}+9+3\sqrt{3}+1}{8}=\frac{6\sqrt{3}+10}{8}
=\frac{2(3\sqrt{3}+5)}{8}
=\frac{3\sqrt{3}+5}{4}
4x3+2x2-8x+7
=4\left(\frac{3\sqrt{3}+5}{4}\right)+2\left(\frac{2+\sqrt{3}}{2}\right)-8\left(\frac{\sqrt{3}+1}{2}\right)+7
=3√3+5+2+√3-4√3-4+7
=4√3+10-4√3
=10
(v) यदि a=\frac{\sqrt{3}+1}{\sqrt{3}-1} और b=\frac{\sqrt{3}-1}{\sqrt{3}-1} तो a2-b2+ab का मान ज्ञात करे ।
Sol :
a2+ab-b2=a2-b2+ab
=(a+b)(a-b)+ab
..........
यदि a और b दो परिेमेय संख्याएँ है तो निम्नलिखित समताओ मे a और b का मान ज्ञात करे :
(i) \frac{3+\sqrt{7}}{3-\sqrt{7}}=a+b\sqrt{7}
Sol :
=\frac{3+\sqrt{7}}{3-\sqrt{7}}\times \frac{3+\sqrt{7}}{3+\sqrt{7}}=a+b\sqrt{7}
\frac{9+3\sqrt{7}+3\sqrt{7}+7}{(3)^2-(\sqrt{7})^2}=a+b\sqrt{7}
\frac{16+6\sqrt{7}}{9-7}=a+b\sqrt{7}
\frac{2(8+3\sqrt{7})}{2}=a+b\sqrt{7}
8+3√7=a+b√7
परिमेय तथा अपरिमेय संख्याओ को अलग-अलग करने पर,
a=8 ,
b√7=3√7
b=\frac{3\sqrt{7}}{\sqrt{7}}
b=3
(ii) \frac{4+2\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}
Sol :
=\frac{4+2\sqrt{5}}{4-3\sqrt{5}}\times \frac{4+3\sqrt{5}}{4+3\sqrt{5}}=a+b\sqrt{5}
\frac{16+12\sqrt{5}+8\sqrt{5}+30}{(4)^2-(3\sqrt{5})^2}=a+b\sqrt{5}
\frac{46+20\sqrt{5}}{16-45}=a+b\sqrt{5}
{46+20\sqrt{5}}{-29}=a+b\sqrt{5}
-\frac{46}{29}-\frac{20}{29}\sqrt{5}=a+b\sqrt{5}
\begin{array}{l|l}a=-\frac{46}{29}& b\sqrt{5}=\frac{-20}{29}\sqrt{5}\\&b=-\frac{20}{29}\end{array}
यदि \frac{5+\sqrt{3}}{7-4\sqrt{3}}=47a+\sqrt{3}b तो a और b का मान ज्ञात करे ।
Sol :
\frac{5+\sqrt{3}}{7-4\sqrt{3}}\times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}=47a+\sqrt{3}b
\frac{35+20\sqrt{3}+7\sqrt{3}+12}{(7)^2-(4\sqrt{3})^2}=47a+\sqrt{3}
\frac{47+27\sqrt{3}}{49-48}=47a+\sqrt{3}b
47+27\sqrt{3}=47a+\sqrt{3}b
\begin{array}{l|l}47a=47&\sqrt{3}b=27\sqrt{3}\\a=\frac{47}{47}=1&b=\frac{27\sqrt{3}}{\sqrt{3}}\\&b=27\end{array}
निम्नलिखित समताओ मे परिमेय संख्याएँ a और b का मान ज्ञात करे ।
(i) \frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1}=a+b\sqrt{5}
Sol :
=\frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1}=a+b\sqrt{5}
=\frac{(\sqrt{5}-1)(\sqrt{5}-1)+(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}=a+b\sqrt{5}
=\frac{5-\sqrt{5}-\sqrt{5}+1+5+\sqrt{5}+\sqrt{5}+1}{(\sqrt{5})^2-1^2}=a+b\sqrt{5}
\frac{12}{5-1}=a+b\sqrt{5}
\frac{12}{4}=a+b\sqrt{5}
\begin{array}{l|l}a=3&b\sqrt{5}=0\\&b=\frac{0}{\sqrt{5}}=0\end{array}
(ii) \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7+\sqrt{5}}{7-\sqrt{5}}=a+7\sqrt{5}b
Sol :
निम्नलिखित का मान एक धन पूर्णाक के घात के रुप मे लिखे:
(i) 73.93
Sol : 633
(ii) 7-3.(9)-3
Sol : 63-3
=\frac{1}{(63)^3}
(iii) 172.175
Sol :
=172+5
=177
(iv) 172.17-5
=172-5
=17-3
(v) (52)7
=514
(vi) (52)-7
=5-14
(vii) \frac{23^{10}}{23^{7}}
Sol :
=2310-7
=233
(viii) \frac{(23)^{-10}}{(23)^7}
Sol :
=23-10-7
=23-17
=\frac{1}{(23)^{17}}
सरल कीजिए:
(i) 2^{\frac{2}{3}}.2^{\frac{1}{3}}
Sol :
=2^{\frac{2}{3}+\frac{1}{3}}
=2^{\frac{2+1}{3}}
=2^{\frac{3}{3}}=2
(ii) \left(3^{\frac{1}{5}}\right)^4
Sol :
=3^{\frac{1}{5}\times 4}
=3^{\frac{4}{5}}
(iii) 13^{\frac{1}{5}.17^{\frac{1}{5}}}
Sol :
=(13\times 17)^{\frac{1}{5}}
=(221)^{\frac{1}{5}}
(iv) \dfrac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}
Sol :
=7^{\frac{1}{5}-\frac{1}{3}}
=7^{\frac{3-5}{15}}
=7^{\frac{-2}{15}}=\frac{1}{7^{\frac{2}{15}}}
निम्नलिखित को सरल करे:
(i) √15×√7
Sol :
√15×7
=√105
(ii) ∛18×∛15
Sol :
=∛18×15
=∛2×3×3×3×5
=3∛10
(iii) ∜5×∜8
Sol :
=∜5×8
=∜40
=∜5×2×2×2×2×\frac{1}{2}
=2\sqrt[4]{\frac{5}{2}}
(iv) \sqrt[7]{9}\times \sqrt[7]{5}\times \sqrt[7]{2}
Sol :
=\sqrt[7]{9\times 5 \times 2}
=\sqrt[7]{90}
(v) \sqrt[8]{12} \div \sqrt[8]{3}
Sol :
=\frac{\sqrt[8]{12}}{\sqrt[8]{3}}
=\sqrt[8]{\frac{12}{3}}
=\sqrt[8]{4}
(vi) \sqrt[5]{24} \div \sqrt[5]{6}
Sol :
=\frac{\sqrt[5]{24}}{\sqrt[5]{6}}
=\sqrt[5]{\frac{24}{6}}
=\sqrt[5]{4}
निम्नलिखित को सरल करे:
(i) ∛2×√5
Sol :
=2^{\frac{1}{3}}\times 5^{\frac{1}{2}}
=2^{\frac{1}{3}\times \frac{1}{2}\times 2}\times 5^{\frac{1}{2}\times \frac{1}{3}\times 3}
=(2^2)^{\frac{1}{6}}\times (5^{3})^{\frac{1}{6}}
=(4)^{\frac{1}{6}}\times(125)^{\frac{1}{6}}
=(500)^{\frac{1}{6}}
=\sqrt[6]{500}
(ii) ∛7×√2
Sol :
(iii) ∛5×√3
Sol :
(iv) ∛7×∜3
Sol :
=7^{\frac{1}{3}} \times 3^{\frac{1}{4}}
=7^{\frac{1}{3}\times \frac{1}{4}\times 4} \times 3^{\frac{1}{4}\times \frac{1}{3}\times 3}
=(7^{4})^{\frac{1}{12}}\times (3^3)^{\frac{1}{12}}
=(64827)^{\frac{1}{12}}
=\sqrt[12]{64827}
(v) √2.∛3.∜4
Sol :
=2^{\frac{1}{2}}\times 3^{\frac{1}{3}}\times 4^{\frac{1}{4}}
=2^{\frac{1}{2}}\times 3^{\frac{1}{3}}\times 4^{\frac{1}{4}}
=2^{\frac{1}{2}\times \frac{1}{6}\times 6}\times 3^{\frac{1}{3}\times \frac{1}{4}\times 4}\times 4^{\frac{1}{4}\times \frac{1}{3}\times 3}
=(2^6)^{\frac{1}{12}}\times (3^4)^{\frac{1}{12}}\times (4^{3})^{\frac{1}{12}}
=(64)^{\frac{1}{12}}\times (81)^{\frac{1}{12}}\times (64)^{\frac{1}{12}}
=(331776)^{\frac{1}{12}}
=\sqrt[12]{331776}
(vi) √3.∛4.∜5
Sol :
(vii) 24÷ ∛200
Sol :
\frac{24}{\sqrt[3]{200}}
=\frac{24}{\sqrt[3]{\frac{2\times 2\times 2\times 5\times 5}{12}}}
=\frac{24}{2\times \sqrt[3]{25}}
=\frac{12}{\sqrt[3]{25}}
=\frac{12}{25^{\frac{1}{3}}}
=\frac{12^{frac{1}{3}\times 3}}{25^{\frac{1}{3}}}
=\frac{(12^3)^{\frac{1}{3}}}{25^{\frac{1}{3}}}
=\left(\frac{1728}{25}\right)^{\frac{1}{3}}
=\sqrt[3]{\frac{1728}{25}}
(viii) ∜36÷∛6
Sol :
=\frac{\sqrt[4]{36}}{\sqrt[4]{6}}
=\frac{(36)^{\frac{1}{4}}}{6^{\frac{1}{3}}}
=\dfrac{6^{2\times \frac{1}{4}}}{6^{\frac{1}{3}}}
=\frac{6^{\frac{1}{6}}}{6^{\frac{1}{3}}}
=6^{\frac{1}{2}-\frac{1}{3}}
=6^{\frac{3-2}{6}}
=6^{\frac{1}{6}}
यदि √2=1.414 , √3=1.732 , √5=2.236 तथा √10=3.162 तो निम्नलिखित के मान निकाले:
(i) \frac{1}{\sqrt{2}}
Sol :
=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}
=\frac{\sqrt{2}}{2}=\frac{1.414}{2}
=0.707
(ii) \frac{3}{\sqrt{10}}
Sol :
(iii) \frac{2+\sqrt{3}}{3}
Sol :
(iv) \frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}
Sol :
यदि m औऱ n दो प्राकृत संख्याएँ हो ताकि mn=25 तो nm का मान होगा:
(i) 4
(ii) 10
(iii) 32
(iv) 16
Sol :
⇒mn=25
⇒mn=52
⇒nm=25=32
\sqrt{10}\times \sqrt{15} बराबर है
(i) 5√6
(ii) 6√5
(iii) √30
(iv) √25
Sol :
=√10×√15
=√2×5×3×5
=5√6
\sqrt[5]{6}\times \sqrt[5]{6^{0}} बराबर है
(i) \sqrt[5]{36}
(ii) \sqrt[5]{6\times 0}
(iii) \sqrt[5]{6}
(iv) \sqrt[5]{12}
Sol :
=\sqrt[5]{6}\times \sqrt[5]{6^0}
=\sqrt[5]{6}
\sqrt[3]{8^2} बराबर है
(i) 8^{\frac{2}{3}}
(ii) 8^{\frac{3}{2}}
(iii) 4\times 4^{\frac{2}{3}}
(iv) 4
Sol :
=\sqrt[3]{8^2}=8^{2\times \frac{1}{3}}
=8^{\frac{2}{3}}
=2^{3\times \frac{2}{3}}
=2^2=4
सिद्ध करे कि √2+√3 एक अपरिमेय संख्या है
Sol :
माना √2+√3 एक परिमेय संख्या है ।
\sqrt{2}+\sqrt{3}=\frac{a}{b} (जहाँ a और b पूर्णांक है ।)
दोनो तरफ वर्ग करने पर,
(\sqrt{2}+\sqrt{3})^2=\frac{a^2}{b^2}
(\sqrt{2})^2+(\sqrt{3})^2+2.\sqrt{2}.\sqrt{3}=\frac{a^2}{b^2}
2\sqrt{6}=\frac{a^2}{b^2}-5
2\sqrt{6}=\frac{a^2-5b^2}{b^2}
\sqrt{6}=\frac{a^2-5b^2}{2b^2}
R.H.S मे a,b, 2 तथा 5 एक पूर्णांक है ।
∴\frac{a^2-5b^2}{2b^2} एक परिमेय संख्या है ।
√6 एक परिमेय होगा ।
लेकिन हमारी अवधारणा गलत की √6 एक परिमेय संख्या है।
अत: √2+√3 एक अपरिमेय संख्या है
परिमेय
Question 3
निम्नलिखित समीकरणो में किसका हल अपरिमेंय संख्याओ को निरुपित करते है ?(i) x2=5
Sol :
⇒x=±√5
अपरिमेय
(ii) x^2=\dfrac{16}{9}
Sol :
⇒x=\sqrt{\dfrac{16}{9}}
⇒x=\pm \dfrac{4}{3}
परिमेय
(iii) (x-1)^2=\dfrac{49}{16}
Sol :
⇒x-1=\sqrt{\dfrac{49}{16}}
⇒x-1=\pm \dfrac{7}{4}
⇒x-1= +\dfrac{7}{4} and x-1=-\dfrac{7}{4}
⇒x=\dfrac{7}{4}+1 and x=-\dfrac{7}{4}+1
⇒x=\dfrac{7+4}{4} and x=\dfrac{-7+4}{4}
⇒x=\dfrac{11}{4} and x=\dfrac{-3}{4}
परिमेय
(iv) (x+1)(x-1)=0
Sol :
⇒x2-1=0
⇒x=±√1
⇒x=-1,+1
परिमेय
(v) x^2=\dfrac{19}{29}
Sol :
⇒x=\sqrt{\dfrac{19}{29}}
अपरिमेय
(vi) (x-1)=5
Sol :
⇒x=5+1
⇒x=6
(i) योगफल परिमेय संख्या है ।
Sol :
-√3+√3=0
-√3,√3
(vi) (x-1)=5
Sol :
⇒x=5+1
⇒x=6
परिमेय
Question 4
प्रत्येक के लिए दो अपरिमेय संख्याओ का उदाहरण दे जिससे उनका:(i) योगफल परिमेय संख्या है ।
Sol :
-√3+√3=0
-√3,√3
(ii) योगफल अपरिमेय संख्या है ।
Sol :
Sol :
2√3+√5=0
2√3,√5(iii) अंतर परिमेय संख्या है ।
Sol :
√3-√3=0
√3,√3
(iv) अंतर अपरिमेय संख्या है ।
√3-√3=0
√3,√3
(iv) अंतर अपरिमेय संख्या है ।
Sol :
2√3-√3=√3
2√3,√3
(v) गुणनफल परिमेय संख्या है ।
Sol :2√3-√3=√3
2√3,√3
(v) गुणनफल परिमेय संख्या है ।
2√3×√3=6
2√3,√3
(vi) गुणनफल अपरिमेय संख्या है ।
Sol :
2√3×2√2=4√6
2√3,2√2
(vii) भागफल परिमेय संख्या है ।
Sol :
\frac{2\sqrt{3}}{\sqrt3}
2√3,√3
(viii) भागफल अपरिमेय संख्या है ।
Sol :
\frac{4\sqrt{6}}{\sqrt{5}}
4√6,√5
Question 5
एक परिमेय एवं अपरिमेय संख्या का उदाहरण दे जिनका गुणनफल परिमेय संख्या होता है ।Sol :
⇒0 , 2√3
परिमेय , अपरिमेय
⇒0×2√3=0 परिमेय
Question 6
निम्नलिखित व्यंजको मे से प्रत्येक को सरल करे :(i) (5+√5)(5-√5)
Sol :
⇒(5)2-(√5)2
⇒25-5
⇒20
Sol :
⇒5(2+√5)+√7(2+√5)
⇒10+5√5+2√7+√35
(iii) (√11-√7)(√11+√7)
Sol :
Using identity:
(a+b)(a-b)=a2-b2
⇒(√11)2-(√7)2
⇒11-7
⇒4
Sol :
Using identity:
(a+b)(a-b)=a2-b2
⇒(√11)2-(√7)2
⇒11-7
⇒4
(iv) (11+√11)(11-√11)
Sol :
⇒11(11-√11)+√11(11-√11)
⇒121-11√11+11√11-11
⇒110
Sol :
⇒11(11-√11)+√11(11-√11)
⇒121-11√11+11√11-11
⇒110
(v) (3+√2)(3-√2)
Sol :
⇒3(3-√2)+√2(3-√2)
⇒9-3√2+3√2-2
⇒7
Sol :
⇒3(3-√2)+√2(3-√2)
⇒9-3√2+3√2-2
⇒7
(vi) (√3+√7)2
Sol :
⇒(√3)2+(√7)2+2(√3)(√7)
⇒3+7+2√21
⇒10+2√21
Sol :
⇒(√3)2+(√7)2+2(√3)(√7)
⇒3+7+2√21
⇒10+2√21
Question 7
निम्नलिखित को सरल करे:
(i) 5√2+4√2
Sol :
⇒√2(5+4)
⇒9√2
(ii) 3√7+2√7
Sol :
⇒√7(3+2)
⇒5√7
(iii) 8√3-5√3
Sol :
⇒√3(8-5)
⇒3√3
(iv) 4√7+5√7-3√7
Sol :
⇒√7(4+5-3)
⇒6√7
(v) 8\sqrt[3]{5}+7\sqrt[3]{5}-13\sqrt[3]{5}
Sol :
⇒\sqrt[3]{5}(8+7-13)
⇒2\sqrt[3]{5}
(vi) 5√3+2√27
Sol :
⇒5√3+2√3×3×3
⇒ 5√3+2×3√3
⇒ 5√3+6√3
⇒√3(6+5)
⇒11√3
Sol :
⇒√2(5+4)
⇒9√2
(ii) 3√7+2√7
Sol :
⇒√7(3+2)
⇒5√7
(iii) 8√3-5√3
Sol :
⇒√3(8-5)
⇒3√3
(iv) 4√7+5√7-3√7
Sol :
⇒√7(4+5-3)
⇒6√7
(v) 8\sqrt[3]{5}+7\sqrt[3]{5}-13\sqrt[3]{5}
Sol :
⇒\sqrt[3]{5}(8+7-13)
⇒2\sqrt[3]{5}
(vi) 5√3+2√27
Sol :
⇒5√3+2√3×3×3
⇒ 5√3+2×3√3
⇒ 5√3+6√3
⇒√3(6+5)
⇒11√3
Question 8
निम्नलिखित को सरल करे:
(i) 4√3-3√2+2√75
Sol :
⇒4√3-3√2+2√5×5×3
⇒4√3-3√2+10√3
⇒4√3+10√3-3√2
⇒√3(4+10)-3√2
⇒14\sqrt{3}-3\sqrt{2}
(ii) √8+√32-√2
Sol :
⇒ √2×2×2+√2×2×2×2×2-√2
⇒2√2+4√2-√2
⇒2√2+3√2
⇒5√2
(iii) \sqrt{192}-\dfrac{1}{2}\sqrt{48}-\sqrt{75}
Sol :
⇒\sqrt{8\times 8\times 3}-\dfrac{1}{2}\sqrt{4\times 4\times 3}-\sqrt{5\times 5\times 3}
⇒8\sqrt{3}-\dfrac{1}{2}\times 4\sqrt{3}-5\sqrt{3}
⇒\sqrt{3} \left(8-\dfrac{4}{2}-5\right)
⇒√3(8-2-5)
⇒√3
Sol :
⇒4√3-3√2+2√5×5×3
⇒4√3-3√2+10√3
⇒4√3+10√3-3√2
⇒√3(4+10)-3√2
⇒14\sqrt{3}-3\sqrt{2}
(ii) √8+√32-√2
Sol :
⇒ √2×2×2+√2×2×2×2×2-√2
⇒2√2+4√2-√2
⇒2√2+3√2
⇒5√2
(iii) \sqrt{192}-\dfrac{1}{2}\sqrt{48}-\sqrt{75}
Sol :
⇒\sqrt{8\times 8\times 3}-\dfrac{1}{2}\sqrt{4\times 4\times 3}-\sqrt{5\times 5\times 3}
⇒8\sqrt{3}-\dfrac{1}{2}\times 4\sqrt{3}-5\sqrt{3}
⇒\sqrt{3} \left(8-\dfrac{4}{2}-5\right)
⇒√3(8-2-5)
⇒√3
Question 9
निम्नलिखित के मान ज्ञात करे :
(i) (2√2+5√3)+(√2-3√3)
(i) (2√2+5√3)+(√2-3√3)
Sol :
⇒2√2+5√3+√2-3√3
⇒3√2+2√3
(ii) 6√5×2√5
⇒2√2+5√3+√2-3√3
⇒3√2+2√3
(ii) 6√5×2√5
Sol :
⇒12×5
⇒60
(iii) 8√15÷2√3
⇒12×5
⇒60
(iii) 8√15÷2√3
Sol :
⇒\frac{8\sqrt{15}}{2\sqrt3}
⇒\frac{8\sqrt{15}}{2\sqrt3}
⇒4√5
Question 10
(i) 5√2
Sol :
√2
(ii) 2√2
Sol :
√2
(iii) √7
Sol :
√7
(iv) √15
Sol :
√15
Question 11
Sol :
(i) \sqrt[5]{a^3b^2c}
(ii) \sqrt[9]{a^7b^5c}
Question 12
(i) \frac{1}{\sqrt2}
Sol :
हर का परिमेयकरण करने पर,
\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}
(ii) \frac{1}{\sqrt{12}}
Sol :
\frac{1}{\sqrt{12}}=\frac{1}{\sqrt{2\times 2 \times 3}}
=\frac{1}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
=\frac{\sqrt{3}}{2\times 3}=\frac{\sqrt{3}}{6}
(iii) \frac{2\sqrt{7}}{\sqrt{11}}
Sol :
\frac{2\sqrt{7}}{\sqrt{11}}\times \frac{\sqrt{11}}{\sqrt{11}}
=\frac{2\sqrt{77}}{11}
(iv) \frac{2}{\sqrt{17}}
Sol :
Question 13
(i) \frac{1}{\sqrt{2}+1}=\dots
Sol :
हर का परिमेयकरण करने पर,
\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}
=\frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}
=\frac{\sqrt{2}-1}{2-1}=\frac{\sqrt{2}-1}{1}
=\sqrt{2}-1
(ii) \frac{1}{2-\sqrt{3}}=\dots
Sol :
(iii) \frac{3}{\sqrt{5}+\sqrt{3}}=\dots
Sol :
(iv) \frac{7}{\sqrt{5}-\sqrt{3}}=\dots
Sol :
Question 14
(i) √48
Sol :
=√2×2×2×2×3
=2×2√3
=4√3
(ii) √175
Sol :
(iii) ∛72
Sol :
=∛2×2×2×3×3
=2×∛9 या
=∛2×2×2×3×3×3×\frac{1}{3}
=6\sqrt[3]{\frac{1}{3}}
(iv) √125
Sol :
(v) ∛54
Sol :
(vi) ∛144
Sol :
(vii) \sqrt[5]{320}
Sol :
(viii) \sqrt{\frac{125}{63}}
Sol :
(ix) \sqrt{\frac{112}{45}}
Sol :
=\sqrt{\frac{2\times 2 \times2 \times 2\times 7 }{3\times 3 \times 5}}
=\frac{4}{3}\sqrt{\frac{7}{5}}
Question 15
(i) \frac{1}{2+\sqrt{3}}
Sol :
(ii) \frac{1}{7+3\sqrt{2}}
Sol :
=\frac{1}{7+3\sqrt{2}}\times \frac{7-3\sqrt{2}}{7-3\sqrt{2}}
=\frac{7-3\sqrt{2}}{(7)^2-(3\sqrt{2})^2}
=\frac{7-3\sqrt{2}}{49-18}
=\frac{7-3\sqrt{2}}{31}
(iii) \frac{5}{\sqrt{3}-\sqrt{5}}
(iv) \frac{6}{3\sqrt{2}-2\sqrt{3}}
Sol :
=\frac{6}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^2-(2\sqrt{3})^2}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{18-12}
=\frac{6(3\sqrt{2}+2\sqrt{3})}{6}
=3√2+2√3
(v) \frac{4}{\sqrt{5}+\sqrt{3}}
Sol :
(vi) \frac{5+\sqrt{6}}{5-\sqrt{6}}
Sol :
=\frac{5+\sqrt{6}}{5-\sqrt{6}}\times \frac{5+\sqrt{6}}{5+\sqrt{6}}
=\frac{(5+\sqrt{6})(5+\sqrt{6}}{(5)^2-(\sqrt{6}^2}
=\frac{25+5\sqrt{6}+5\sqrt{6}+6}{25-6}
=\frac{31+10\sqrt{6}}{19}
(vii) \frac{3+\sqrt{2}}{3-\sqrt{2}}
(viii) \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}
Sol :
=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times \frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}
=\frac{6\sqrt{30}+4\times 6-3\times 5-2\sqrt{30}}{45-24}
=\frac{4\sqrt{30}+19}{21}
(ix) \frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}
Sol :
Question 16
(i) \frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}
Sol :
हर का परिमेय करण करने पर,
=\frac{3}{5-\sqrt{3}}\times \frac{5+\sqrt{3}}{5+\sqrt{3}}+\frac{2}{5+\sqrt{3}}\times \frac{5-\sqrt{3}}{5-\sqrt{3}}
=\frac{3(5+\sqrt{3})}{(5)^2-(\sqrt{3})^2}+\frac{2(5-\sqrt{3})}{(5)^2-(\sqrt{3})^2}
=\frac{3(5+\sqrt{3})}{25-3}+\frac{2(5-\sqrt{3})}{25-3}
=\frac{3(5+\sqrt{3})}{22}+\frac{2(5-\sqrt{3})}{22}
=\frac{3(5+\sqrt{3}+2(5-\sqrt{3}))}{22}
=\frac{15+3\sqrt{3}+10-2\sqrt{3}}{22}
=\frac{25+\sqrt{3}}{22}
(ii) \frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}
Sol :
=\frac{(\sqrt{5}-2)(\sqrt{5}-2)-(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}+2)(\sqrt{5}-2)}
=\frac{(5-2\sqrt{5}-2\sqrt{5}+4)-(5+2\sqrt{5}+2\sqrt{5}+4)}{(\sqrt{5})^2-(2)^2}
=\frac{(9-4\sqrt{5})-(9+4\sqrt{5})}{5-4}
=\frac{9-4\sqrt{5}-9-4\sqrt{5}}{1}
=-8√5
(iii) \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}
Sol :
=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}
=\frac{5+\sqrt{15}+\sqrt{15}+3+5-\sqrt{15}-\sqrt{15}+3}{(\sqrt{5})^2-(\sqrt{3})^2}
=\frac{16}{5-3}=\frac{16}{2}=8
(iv) \frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}+\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}
Sol :
=\frac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})+(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}
=\frac{(\sqrt{5}-\sqrt{3}+\sqrt{10}-\sqrt{6}+\sqrt{5}+\sqrt{3}-\sqrt{10}-\sqrt{6})()+()()}{(\sqrt{5})^2-(\sqrt{3})^2}
=\frac{2\sqrt{5}-2\sqrt{6}}{5-3}
=\frac{2(\sqrt{6}-\sqrt{6})}{2}
=√5-√6
Question 17
(i) \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\sqrt{35}
Sol :
हर का परिमेयकरण करने पर,
=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\sqrt{35}
=\frac{7-\sqrt{35}-\sqrt{35}+5}{(\sqrt{7})^2-(\sqrt{5})^2}+\sqrt{35}
=\frac{12-2\sqrt{35}}{7-5}+\sqrt{35}
=\frac{2(6-\sqrt{35})}{2}+\sqrt{35}
=6-√35+√35
=6
(ii) \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+2\sqrt{6}
Sol :
हर का परिमेयकरण करने पर,
=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+2\sqrt{6}
=\frac{3-\sqrt{6}-\sqrt{6}+2}{(\sqrt{3})^2-(\sqrt{2})^2}+2\sqrt{6}
=\frac{5-2\sqrt{3}}{3-2}+2\sqrt{6}
=5-2√6+2√6
=5
Question 18
Sol :
a=3+√8
\frac{1}{a}=\frac{1}{3+\sqrt{8}}
=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}
=\frac{3-\sqrt{8}}{3^2-(\sqrt{8})^2}
=\frac{3-\sqrt{8}}{9-8}
\frac{1}{a}=3-\sqrt{3}
a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2.a.\frac{1}{a}
=(3+\sqrt{8}+3-\sqrt{8})^2-2
=(6)2-2
=36-2
=34
(ii) यदि a=\frac{\sqrt{2}+1}{\sqrt{2}-1}, b=\frac{\sqrt{2}-1}{\sqrt{2}+1} , सिद्ध करे कि a2+b2+ab=35
Sol :
a+b=\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}
=\frac{(\sqrt{2}+1)(\sqrt{2}+1)+(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}
=\frac{2+\sqrt{2}+\sqrt{2}+1+2-\sqrt{2}-\sqrt{2}+1}{(\sqrt{2})^2-(1)^2}
=\frac{6}{2-1}=6
L.H.S
a2+b2+ab=a2+b2+ab
=(a+b)2-2ab+ab
=(a+b)2-ab
=6^2-\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}-1}{\sqrt{2}+1}
=36-1
=35
(iii) यदि a=2+√3 तो a^3+\frac{1}{a^3} का मान ज्ञात करे।
Sol :
a=2+√3
\frac{1}{a}=\frac{1}{2+\sqrt{3}}
=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}
=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}
\frac{1}{a}=2-\sqrt{3}
a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2-a.\frac{1}{a}+\frac{1}{a^2}\right)
=\left(a+\frac{1}{a}\right)\left((a+\frac{1}{a})^2\right)-2a.\frac{1}{a}-1
=4\left[(4)^2-3\right]
=4[16-3]
=4×13
=52
(iv) यदि x=\frac{\sqrt{3}+1}{2} तो 4x3+2x2-8x+7 का मान ज्ञात करे ।
Sol :
x^2=\left(\frac{\sqrt{3}+1}{2}\right)^2
=\frac{(\sqrt{3}^2+2.\sqrt{3}.1+1^2)}{4}
=\frac{4+2\sqrt{3}}{4}=\frac{2(2+\sqrt{3})}{4}
x^2=\frac{2+\sqrt{3}}{2}
x^3=\left(\frac{\sqrt{3}+1}{2}\right)^2
=\frac{(\sqrt{3})^3+3(\sqrt{3})^2.1+3\sqrt{3}.1^2+1^3}{8}
=\frac{3\sqrt{3}+9+3\sqrt{3}+1}{8}=\frac{6\sqrt{3}+10}{8}
=\frac{2(3\sqrt{3}+5)}{8}
=\frac{3\sqrt{3}+5}{4}
4x3+2x2-8x+7
=4\left(\frac{3\sqrt{3}+5}{4}\right)+2\left(\frac{2+\sqrt{3}}{2}\right)-8\left(\frac{\sqrt{3}+1}{2}\right)+7
=3√3+5+2+√3-4√3-4+7
=4√3+10-4√3
=10
(v) यदि a=\frac{\sqrt{3}+1}{\sqrt{3}-1} और b=\frac{\sqrt{3}-1}{\sqrt{3}-1} तो a2-b2+ab का मान ज्ञात करे ।
Sol :
a2+ab-b2=a2-b2+ab
=(a+b)(a-b)+ab
..........
Question 19
(i) \frac{3+\sqrt{7}}{3-\sqrt{7}}=a+b\sqrt{7}
Sol :
=\frac{3+\sqrt{7}}{3-\sqrt{7}}\times \frac{3+\sqrt{7}}{3+\sqrt{7}}=a+b\sqrt{7}
\frac{9+3\sqrt{7}+3\sqrt{7}+7}{(3)^2-(\sqrt{7})^2}=a+b\sqrt{7}
\frac{16+6\sqrt{7}}{9-7}=a+b\sqrt{7}
\frac{2(8+3\sqrt{7})}{2}=a+b\sqrt{7}
8+3√7=a+b√7
परिमेय तथा अपरिमेय संख्याओ को अलग-अलग करने पर,
a=8 ,
b√7=3√7
b=\frac{3\sqrt{7}}{\sqrt{7}}
b=3
(ii) \frac{4+2\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}
Sol :
=\frac{4+2\sqrt{5}}{4-3\sqrt{5}}\times \frac{4+3\sqrt{5}}{4+3\sqrt{5}}=a+b\sqrt{5}
\frac{16+12\sqrt{5}+8\sqrt{5}+30}{(4)^2-(3\sqrt{5})^2}=a+b\sqrt{5}
\frac{46+20\sqrt{5}}{16-45}=a+b\sqrt{5}
{46+20\sqrt{5}}{-29}=a+b\sqrt{5}
-\frac{46}{29}-\frac{20}{29}\sqrt{5}=a+b\sqrt{5}
\begin{array}{l|l}a=-\frac{46}{29}& b\sqrt{5}=\frac{-20}{29}\sqrt{5}\\&b=-\frac{20}{29}\end{array}
Question 20
Sol :
\frac{5+\sqrt{3}}{7-4\sqrt{3}}\times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}=47a+\sqrt{3}b
\frac{35+20\sqrt{3}+7\sqrt{3}+12}{(7)^2-(4\sqrt{3})^2}=47a+\sqrt{3}
\frac{47+27\sqrt{3}}{49-48}=47a+\sqrt{3}b
47+27\sqrt{3}=47a+\sqrt{3}b
\begin{array}{l|l}47a=47&\sqrt{3}b=27\sqrt{3}\\a=\frac{47}{47}=1&b=\frac{27\sqrt{3}}{\sqrt{3}}\\&b=27\end{array}
Question 21
(i) \frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1}=a+b\sqrt{5}
Sol :
=\frac{\sqrt{5}-1}{\sqrt{5}+1}+\frac{\sqrt{5}+1}{\sqrt{5}-1}=a+b\sqrt{5}
=\frac{(\sqrt{5}-1)(\sqrt{5}-1)+(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}=a+b\sqrt{5}
=\frac{5-\sqrt{5}-\sqrt{5}+1+5+\sqrt{5}+\sqrt{5}+1}{(\sqrt{5})^2-1^2}=a+b\sqrt{5}
\frac{12}{5-1}=a+b\sqrt{5}
\frac{12}{4}=a+b\sqrt{5}
\begin{array}{l|l}a=3&b\sqrt{5}=0\\&b=\frac{0}{\sqrt{5}}=0\end{array}
(ii) \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7+\sqrt{5}}{7-\sqrt{5}}=a+7\sqrt{5}b
Sol :
Question 22
(i) 73.93
Sol : 633
(ii) 7-3.(9)-3
Sol : 63-3
=\frac{1}{(63)^3}
(iii) 172.175
Sol :
=172+5
=177
(iv) 172.17-5
=172-5
=17-3
(v) (52)7
=514
(vi) (52)-7
=5-14
(vii) \frac{23^{10}}{23^{7}}
Sol :
=2310-7
=233
Sol :
=23-10-7
=23-17
=\frac{1}{(23)^{17}}
Question 23
(i) 2^{\frac{2}{3}}.2^{\frac{1}{3}}
Sol :
=2^{\frac{2}{3}+\frac{1}{3}}
=2^{\frac{2+1}{3}}
=2^{\frac{3}{3}}=2
(ii) \left(3^{\frac{1}{5}}\right)^4
Sol :
=3^{\frac{1}{5}\times 4}
=3^{\frac{4}{5}}
(iii) 13^{\frac{1}{5}.17^{\frac{1}{5}}}
Sol :
=(13\times 17)^{\frac{1}{5}}
=(221)^{\frac{1}{5}}
(iv) \dfrac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}
Sol :
=7^{\frac{1}{5}-\frac{1}{3}}
=7^{\frac{3-5}{15}}
=7^{\frac{-2}{15}}=\frac{1}{7^{\frac{2}{15}}}
Question 24
(i) √15×√7
Sol :
√15×7
=√105
(ii) ∛18×∛15
Sol :
=∛18×15
=∛2×3×3×3×5
=3∛10
(iii) ∜5×∜8
Sol :
=∜5×8
=∜40
=∜5×2×2×2×2×\frac{1}{2}
=2\sqrt[4]{\frac{5}{2}}
(iv) \sqrt[7]{9}\times \sqrt[7]{5}\times \sqrt[7]{2}
Sol :
=\sqrt[7]{9\times 5 \times 2}
=\sqrt[7]{90}
(v) \sqrt[8]{12} \div \sqrt[8]{3}
Sol :
=\frac{\sqrt[8]{12}}{\sqrt[8]{3}}
=\sqrt[8]{\frac{12}{3}}
=\sqrt[8]{4}
(vi) \sqrt[5]{24} \div \sqrt[5]{6}
Sol :
=\frac{\sqrt[5]{24}}{\sqrt[5]{6}}
=\sqrt[5]{\frac{24}{6}}
=\sqrt[5]{4}
Question 25
(i) ∛2×√5
Sol :
=2^{\frac{1}{3}}\times 5^{\frac{1}{2}}
=2^{\frac{1}{3}\times \frac{1}{2}\times 2}\times 5^{\frac{1}{2}\times \frac{1}{3}\times 3}
=(2^2)^{\frac{1}{6}}\times (5^{3})^{\frac{1}{6}}
=(4)^{\frac{1}{6}}\times(125)^{\frac{1}{6}}
=(500)^{\frac{1}{6}}
=\sqrt[6]{500}
(ii) ∛7×√2
Sol :
(iii) ∛5×√3
Sol :
(iv) ∛7×∜3
Sol :
=7^{\frac{1}{3}} \times 3^{\frac{1}{4}}
=7^{\frac{1}{3}\times \frac{1}{4}\times 4} \times 3^{\frac{1}{4}\times \frac{1}{3}\times 3}
=(7^{4})^{\frac{1}{12}}\times (3^3)^{\frac{1}{12}}
=(64827)^{\frac{1}{12}}
=\sqrt[12]{64827}
(v) √2.∛3.∜4
Sol :
=2^{\frac{1}{2}}\times 3^{\frac{1}{3}}\times 4^{\frac{1}{4}}
=2^{\frac{1}{2}}\times 3^{\frac{1}{3}}\times 4^{\frac{1}{4}}
=2^{\frac{1}{2}\times \frac{1}{6}\times 6}\times 3^{\frac{1}{3}\times \frac{1}{4}\times 4}\times 4^{\frac{1}{4}\times \frac{1}{3}\times 3}
=(2^6)^{\frac{1}{12}}\times (3^4)^{\frac{1}{12}}\times (4^{3})^{\frac{1}{12}}
=(64)^{\frac{1}{12}}\times (81)^{\frac{1}{12}}\times (64)^{\frac{1}{12}}
=(331776)^{\frac{1}{12}}
=\sqrt[12]{331776}
(vi) √3.∛4.∜5
Sol :
(vii) 24÷ ∛200
Sol :
\frac{24}{\sqrt[3]{200}}
=\frac{24}{\sqrt[3]{\frac{2\times 2\times 2\times 5\times 5}{12}}}
=\frac{24}{2\times \sqrt[3]{25}}
=\frac{12}{\sqrt[3]{25}}
=\frac{12}{25^{\frac{1}{3}}}
=\frac{12^{frac{1}{3}\times 3}}{25^{\frac{1}{3}}}
=\frac{(12^3)^{\frac{1}{3}}}{25^{\frac{1}{3}}}
=\left(\frac{1728}{25}\right)^{\frac{1}{3}}
=\sqrt[3]{\frac{1728}{25}}
(viii) ∜36÷∛6
Sol :
=\frac{\sqrt[4]{36}}{\sqrt[4]{6}}
=\frac{(36)^{\frac{1}{4}}}{6^{\frac{1}{3}}}
=\dfrac{6^{2\times \frac{1}{4}}}{6^{\frac{1}{3}}}
=\frac{6^{\frac{1}{6}}}{6^{\frac{1}{3}}}
=6^{\frac{1}{2}-\frac{1}{3}}
=6^{\frac{3-2}{6}}
=6^{\frac{1}{6}}
Question 26
(i) \frac{1}{\sqrt{2}}
Sol :
=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}
=\frac{\sqrt{2}}{2}=\frac{1.414}{2}
=0.707
(ii) \frac{3}{\sqrt{10}}
Sol :
(iii) \frac{2+\sqrt{3}}{3}
Sol :
(iv) \frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}
Sol :
Question 27
(i) 4
(ii) 10
(iii) 32
(iv) 16
Sol :
⇒mn=25
⇒mn=52
⇒nm=25=32
Question 28
(i) 5√6
(ii) 6√5
(iii) √30
(iv) √25
Sol :
=√10×√15
=√2×5×3×5
=5√6
Question 29
(i) \sqrt[5]{36}
(ii) \sqrt[5]{6\times 0}
(iii) \sqrt[5]{6}
(iv) \sqrt[5]{12}
Sol :
=\sqrt[5]{6}\times \sqrt[5]{6^0}
=\sqrt[5]{6}
Question 30
(i) 8^{\frac{2}{3}}
(ii) 8^{\frac{3}{2}}
(iii) 4\times 4^{\frac{2}{3}}
(iv) 4
Sol :
=\sqrt[3]{8^2}=8^{2\times \frac{1}{3}}
=8^{\frac{2}{3}}
=2^{3\times \frac{2}{3}}
=2^2=4
Question 31
Sol :
माना √2+√3 एक परिमेय संख्या है ।
\sqrt{2}+\sqrt{3}=\frac{a}{b} (जहाँ a और b पूर्णांक है ।)
दोनो तरफ वर्ग करने पर,
(\sqrt{2}+\sqrt{3})^2=\frac{a^2}{b^2}
(\sqrt{2})^2+(\sqrt{3})^2+2.\sqrt{2}.\sqrt{3}=\frac{a^2}{b^2}
2\sqrt{6}=\frac{a^2}{b^2}-5
2\sqrt{6}=\frac{a^2-5b^2}{b^2}
\sqrt{6}=\frac{a^2-5b^2}{2b^2}
R.H.S मे a,b, 2 तथा 5 एक पूर्णांक है ।
∴\frac{a^2-5b^2}{2b^2} एक परिमेय संख्या है ।
√6 एक परिमेय होगा ।
लेकिन हमारी अवधारणा गलत की √6 एक परिमेय संख्या है।
अत: √2+√3 एक अपरिमेय संख्या है
Pure answer
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