Hindi, asked by kaamilkumar123, 5 months ago

बंगले के पीछे लगा पेड़ गिर गया।- वाक्य में रेखांकित पदबंध है:-
संजा पटबंध​

Answers

Answered by Anonymous
2

\huge\mathbb{SOLUTION:-}

\sf\bullet a=3+\sqrt{8}

Now find the Value of

\sf\bigg(\dfrac{1}{a}\bigg)=\dfrac{1}{3+\sqrt{8}}\\ \\ \sf\implies \dfrac{1}{3+\sqrt{8}}\times \dfrac{(3-\sqrt{8})}{(3-\sqrt{8})}\\ \\ \sf\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{(3){}^{2}-(\sqrt{8}){}^{2}}\\ \\ \sf\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{9-8}\\ \\ \sf\implies \dfrac{1}{a}= 3-\sqrt{8}

Now find the value of

\sf\bullet \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg)

\boxed{\sf{\red{(a+b){}^{2}= a{}^{2}+2ab+b{}^{2}}}}

\sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}+2\\ \\ \sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=( 3+\cancel{\sqrt{8}}+3-\cancel{\sqrt{8}}){}^{2}\\ \\ \sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}= (6){}^{2}=9\\ \\ \sf\implies 9=a{}^{2}+\dfrac{1}{a{}^{2}}+2\\ \\ \sf\implies 9-2=a{}^{2}+\dfrac{1}{a{}^{2}}\\ \\ \sf\implies 7=a{}^{2}+\dfrac{1}{a{}^{2}}

\boxed{\sf{a{}^{2}+\dfrac{1}{a{}^{2}}=7}}

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