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Answer 1

Solution !

$\sf \ Simplify \\ \\ \sf \ 1.\ i] \ \dfrac{x^{-9}\ y^2}{x^{-7}\ y^8}\\ \\ \sf ii] (a^{-2}- b^{-2})^{-1}\\ \\ \sf 2] Find\ value \ of \ m\\ \\ \sf (-3)^{m+1} \times (-3)^5= (-3)^7$

$\underline{\bigstar{\it\ \ Question\ 1 \ :-}}$

Formula used !

$\sf\bullet a^m\times a^n= a^{(m+n)}\\ \\ \sf\bullet a^{-n}= \dfrac{1}{a^n}\\ \\ \bullet\sf \dfrac{a^n}{a^{-m}}= a^n\times a^m$

$\sf i]\ \ \ \dfrac{x^{-9}\ y^3}{x^{-7}\ y^8}\\ \\ \longrightarrow\sf x^{-9}\ y^3 \times x^7 \ y^{-8}\\ \\ \longrightarrow\sf x^{-9+7}\times y^{3-8}\\ \\ \longrightarrow\sf x^{-2}\times y^{-5}\\ \\ \longrightarrow\sf \dfrac{1}{x^2 \ y^5}$

$\boxed{\sf {\dfrac{1}{x^2\ y^5}}}$

$\sf ii] \ \ \ \ (a^{-2}-b^{-2})^{-1}\\ \\ \longrightarrow\sf \bigg(\dfrac{1}{a^2}-\dfrac{1}{b^2}\bigg)^{-1}\\ \\ \longrightarrow\sf \bigg(\dfrac{b^2-a^2}{a^2\ b^2}\bigg)^{-1}\\ \\ \longrightarrow\sf \dfrac{a^2\ b^2}{b^2-a^2}$

$\boxed{\sf \dfrac{a^2\ b^2}{b^2-a^2}}$

$\underline{\bigstar{\it\ \ Question\ 2 \ :-}}$

$\sf \ \ (-3)^{m+1} \times (-3)^5= (-3)^7\\ \\ \longrightarrow\sf (-3)^{m+1+5}=(-3)^7\\ \\ \longrightarrow\sf\ (-3)^{m+6}= (-3)^7\\ \\ \sf \ \ Their \ base \ are\ same, \ take \ power \\ \\ \longrightarrow\sf m+6= 7\\ \\ \longrightarrow\sf m= 7-6\\ \\ \longrightarrow\sf m=1$

$\boxed{\sf\ \ m= 1}$

Answer 2

Solution !

$\sf \ Simplify$

$\sf \ 1.\ i] \ \dfrac{x^{-9}\ y^2}{x^{-7}\ y^8}$

$\sf ii] (a^{-2}- b^{-2})^{-1}$

$\sf 2] Find\ value \ of \ m$

$\sf (-3)^{m+1} \times (-3)^5= (-3)^7$

$\underline{\bigstar{\it\ \ Question\ 1 \ :-}}$

Formula used !

$\sf\bullet \: \: a^m\times a^n= a^{(m+n)}$

$\sf\bullet \: \: a^{-n}= \dfrac{1}{a^n}$

$\bullet\sf \: \: \dfrac{a^n}{a^{-m}}= a^n\times a^m$

$\sf i]\ \ \ \dfrac{x^{-9}\ y^3}{x^{-7}\ y^8}$

$\longrightarrow\sf x^{-9}\ y^3 \times x^7 \ y^{-8}$

$\longrightarrow\sf x^{-9+7}\times y^{3-8}$

$\longrightarrow\sf x^{-2}\times y^{-5}$

$\longrightarrow\sf \dfrac{1}{x^2 \ y^5}$

$\boxed{\sf {\dfrac{1}{x^2\ y^5}}}$

$\sf ii] \ \ \ \ (a^{-2}-b^{-2})^{-1}$

$\longrightarrow\sf \bigg(\dfrac{1}{a^2}-\dfrac{1}{b^2}\bigg)^{-1}$

$\longrightarrow\sf \bigg(\dfrac{b^2-a^2}{a^2\ b^2}\bigg)^{-1}$

$\longrightarrow\sf \dfrac{a^2\ b^2}{b^2-a^2}$

$\boxed{\sf \dfrac{a^2\ b^2}{b^2-a^2}}$

$\underline{\bigstar{\it\ \ Question\ 2 \ :-}}$

$\sf \ \ (-3)^{m+1} \times (-3)^5= (-3)^7$

$\longrightarrow\sf (-3)^{m+1+5}=(-3)^7$

$\longrightarrow\sf\ (-3)^{m+6}= (-3)^7$

$\sf \ \ Their \ base \ are\ same, \ take \ power$

$\longrightarrow\sf m+6= 7$

$\longrightarrow\sf m= 7-6$

$\longrightarrow\sf m=1$

$\boxed{\sf\ \ m= 1}$