Question

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Answer 1

Correct question is

If $\mathrm{x=(\frac{a}{b})^{\frac{2ab}{a^{2}-b^{2}}}}$ ,

prove that $\mathrm{\frac{ab}{a^{2}+b^{2}}\lgroup x^{\frac{a}{b}}+x^{\frac{b}{a}}\rgroup=(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}}$

Proof :

Now, $\mathrm{\frac{ab}{a^{2}+b^{2}}\lgroup x^{\frac{a}{b}}+x^{\frac{b}{a}}\rgroup}$

$\mathrm{=\frac{ab}{a^{2}+b^{2}}\lgroup (\frac{a}{b})^{\frac{2a^{2}}{a^{2}-b^{2}}}+(\frac{a}{b})^{\frac{2b^{2}}{a^{2}-b^{2}}}\rgroup}$

$\small{\mathrm{=\frac{ab}{a^{2}-b^{2}}\lgroup (\frac{a}{b})^{\frac{(a^{2}+b^{2})+(a^{2}-b^{2})}{a^{2}-b^{2}}}+(\frac{a}{b})^{\frac{(a^{2}+b^{2})-(a^{2}-b^{2})}{a^{2}-b^{2}}}\rgroup}}$

$\mathrm{=\frac{ab}{a^{2}+b^{2}}(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}\lgroup (\frac{a}{b})^{1}+(\frac{a}{b})^{-1}\rgroup}$

$\mathrm{=\frac{ab}{a^{2}+b^{2}}(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}\lgroup \frac{a}{b}+\frac{b}{a}\rgroup}$

$\mathrm{=\frac{ab}{a^{2}+b^{2}}(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}\frac{a^{2}+b^{2}}{ab}}$

$\mathrm{=(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}}$

$\to \boxed{\mathrm{\frac{ab}{a^{2}+b^{2}}\lgroup x^{\frac{a}{b}}+x^{\frac{b}{a}}\rgroup=(\frac{a}{b})^{\frac{a^{2}+b^{2}}{a^{2}-b^{2}}}}}$

Hence, proved.