пожалуйста помогите. хоть что нибудь

Question 1

Question 2

6)
$( \frac{a^{-1}-b^{-1}}{a^{-3}+b^{-3}}: \frac{a^2b^2}{(a+b)^2-3ab}+1 )^{-1}-0.5= \\ \\ =( \frac{ \frac{1}{a}- \frac{1}{b}}{ \frac{1}{a^3}+ \frac{1}{b^3} }* \frac{a^2+2ab+b^2-3ab}{a^2b^2}+1 )^{-1}-0.5= \\ \\ =( \frac{ \frac{b-a}{ab} }{ \frac{b^3+a^3}{a^3b^3} }* \frac{a^2-ab+b^2}{a^2b^2} +1 )^{-1}-0.5= \\ \\ =( \frac{(b-a)a^3b^3}{ab(a^3+b^3)}* \frac{a^2-ab+b^2}{a^2b^2}+1 )^{-1}-0.5= \\ \\ =( \frac{(b-a)}{(a+b)(a^2-ab+b^2)}* \frac{a^2-ab+b^2}{1} +1 )^{-1}-0.5= \\ \\$
$=( \frac{b-a}{a+b}+1 )^{-1}-0.5=( \frac{b-a+a+b}{a+b} )^{-1}-0.5= \\ \\ =( \frac{2b}{a+b} )^{-1}-0.5= \frac{a+b}{2b}- \frac{1}{2}= \frac{a+b-b}{2b}= \frac{a}{2b}$

7)
$\frac{m^{-2}n^{-1}-m^{-1}n^{-2}}{m^{-2}-n^{-2}}- \frac{1}{m}(mn^{-1}+2+m^{-1}n)^{-1}= \\ \\ = \frac{ \frac{1}{m^2n}- \frac{1}{mn^2} }{ \frac{1}{m^2}- \frac{1}{n^2} }- \frac{1}{m}( \frac{m}{n} +2+ \frac{n}{m} )^{-1} = \\ \\ = \frac{ \frac{n-m}{m^2n^2} }{ \frac{n^2-m^2}{m^2n^2} }- \frac{1}{m}( \frac{m^2+2mn+n^2}{mn} )^{-1}= \\ \\ = \frac{n-m}{m^2n^2}* \frac{m^2n^2}{(n-m)(n+m)}- \frac{1}{m}( \frac{(m+n)^2}{mn} )^{-1}= \\ \\ = \frac{1}{m+n} - \frac{1}{m}* \frac{mn}{(m+n)^2}=$
$= \frac{1}{m+n}- \frac{n}{(m+n)^2}= \frac{m+n-n}{(m+n)^2}= \frac{m}{(m+n)^2}= \frac{0.003}{(0.003+0.007)^2}= \\ \\ = \frac{0.003}{0.01^2}= \frac{0.003}{0.0001}=30$