Question

pls help me to solve this question pls​

Answer 1

$\underline{\text{We have to simplify the given term:}}$

$\therefore\mathrm{\dfrac{\big(a^{\frac{1}{m}}-a^{\frac{1}{n}}\big)^{2}+4a^{\frac{m+n}{mn}}}{\big(a^{\frac{2}{m}}-a^{\frac{2}{n}}\big)\big( \sqrt[m]{a^{m+1}}+ \sqrt[n]{a^{n+1}}\big)}}$

$\quad\boxed{\tiny{\mathsf{Using\:(a-b)^{2}=a^{2}-2ab+b^{2}\:\&\:\sqrt[n]{a}=a^{\frac{1}{n}}}}}:$

$\mathrm{=\dfrac{a^{\frac{2}{m}}+a^{\frac{2}{n}}-2a^{\frac{1}{m}+\frac{1}{n}}+4a^{\frac{1}{m}+\frac{1}{n}}}{\big(a^{\frac{2}{m}}-a^{\frac{2}{n}}\big)\big( a^{\frac{m+1}{m}}+a^{\frac{n+1}{n}}\big)}}$

$\mathrm{=\dfrac{a^{\frac{2}{m}}+2a^{\frac{1}{m}+\frac{1}{n}}+a^{\frac{2}{n}}}{\big( a^{\frac{2}{m}}-a^{\frac{2}{n}}\big)\:a\:\big(a^{\frac{1}{m}}+a^{\frac{1}{n}}\big)}}$

$\quad\boxed{\tiny{\mathsf{Using\:(a+b)^{2}=a^{2}+2ab+b^{2}\:\&\:a^{2}-b^{2}=(a+b)(a-b)}}}:$

$\mathrm{=\dfrac{\big(a^{\frac{1}{m}}+a^{\frac{1}{n}}\big)^{2}}{a\:\big(a^{\frac{1}{m}}+a^{\frac{1}{n}}\big)\big(a^{\frac{1}{m}}-a^{\frac{1}{n}}\big)\big(a^{\frac{1}{m}}+a^{\frac{1}{n}}\big)}}$

$\mathrm{=\dfrac{1}{a\:\big(a^{\frac{1}{m}}-a^{\frac{1}{n}}\big)}}$

$\mathrm{=\dfrac{1}{a\:\big(\sqrt[m]{a}-\sqrt[n]{a}\big)}}$

$\underline{\boxed{\tiny{\bold{\dfrac{\big(a^{\frac{1}{m}}-a^{\frac{1}{n}}\big)^{2}+4a^{\frac{m+n}{mn}}}{\big(a^{\frac{2}{m}}-a^{\frac{2}{n}}\big)\big( \sqrt[m]{a^{m+1}}+ \sqrt[n]{a^{n+1}}\big)}=\dfrac{1}{a\:\big(\sqrt[m]{a}-\sqrt[n]{a}\big)}}}}}$

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