Math, asked by abhiishekkumar226, 8 months ago

solve it step by step​

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Answered by chnageswarr
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Answered by Abhishek474241
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AnSwEr

{\tt{\red{\underline{\large{Given}}}}}

\tt{X+\dfrac{1}{X}}=2

{\sf{\green{\underline{\large{To\:find}}}}}

\tt{X^3+\dfrac{1}{X^3}(2)x^2+\frac{1}{x^2}(3){x^4+\frac{1}{x^4}}

{\sf{\pink{\underline{\Large{Explanation}}}}}

We know that

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

Therefore

\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}

Solving

\tt{X+\dfrac{1}{X}}=2

Both side squaring

\tt{(X+\dfrac{1}{X})^2}=(2)²

\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=4

\implies\tt{4=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}

\implies\tt{4=X^2+\dfrac{1}{X^2}+2}

\implies\tt{4-2=X^2+\dfrac{1}{X^2}}

\implies\tt{2=X^2+\dfrac{1}{X^2}}

Now

\tt{X^3+\dfrac{1}{X^3}}

Formula used

\implies\tt{X^3+\dfrac{1}{X^3)}=(X+\dfrac{1}{x})(X^2+\dfrac{1}{X^2}-\frac{1}{X}\times{X)}}

utting value

\implies\tt{X^3+\dfrac{1}{X^3}={2}(2-1)}

\implies\tt{X^3+\dfrac{1}{X^3}={2}(1)}

\implies\tt{X^3+\dfrac{1}{X^3}={2}}

Now for

\rightarrow\sf{x^4+\frac{1}{x^4}}

\implies\tt{2=X^2+\dfrac{1}{X^2}}

Both side squaring

\rightarrow\sf{x^4+\frac{1}{x^4}}+2=4

\rightarrow\sf{x^4+\frac{1}{x^4}}=4-2

\rightarrow\sf{x^4+\frac{1}{x^4}}=2

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