Пожалуйста помогите решить примеры из номера 566.

Question 1

Question 2

№566
1)
$\frac{ax}{a+x} - \frac{bx}{b-x}= \frac{ax(b-x)-bx(a+x)}{(a+x)(b-x)}= \frac{axb-ax^2-bxa-bx^2}{(a+x)(b-x)}= \frac{-x^2(a+b)}{(a+x)(b-x)}$
$x= \frac{ab}{a-b}$
$\frac{-x^2(a+b)}{(a+x)(b-x)} = \frac{-(\frac{ab}{a-b})^2(a+b)}{(a+\frac{ab}{a-b})(b-\frac{ab}{a-b})} =\frac{-(\frac{ab}{a-b})^2(a+b)}{\frac{a^2-ab+ab}{a-b}*\frac{ab-b^2-ab}{a-b}} =\frac{-(\frac{ab}{a-b})^2(a+b)}{\frac{a^2}{a-b}*(\frac{-b^2}{a-b})} =$ $=\frac{-\frac{(ab)^2}{(a-b)^2}*(a+b)}{-\frac{a^2b^2}{(a-b)^2}} = \frac{a^2b^2*(a+b)}{(a-b)^2}* \frac{(a-b)^2}{a^2b^2}= a+b$
2)
$\frac{x^2y^2}{ x^{2} -y^2}$ при $x= \frac{2ab}{a^2-b^2}$ $y= \frac{2ab}{a^2+b^2}$
$\frac{ (\frac{2ab}{a^2-b^2} )^2*( \frac{2ab}{a^2+b^2})^2 }{( \frac{2ab}{a^2-b^2})^2-(\frac{2ab}{a^2+b^2})^2 }= \frac{ (\frac{2ab}{a^2-b^2} * \frac{2ab}{a^2+b^2})^2 }{( \frac{2ab}{a^2-b^2}-\frac{2ab}{a^2+b^2})(\frac{2ab}{a^2-b^2}+\frac{2ab}{a^2+b^2} ) }=$ $= \frac{ (\frac{4a^2b^2}{(a^2-b^2)(a^2+b^2)} )^2 }{\frac{2ab(a^2+b^2-a^2+b^2)}{(a^2-b^2)(a^2+b^2)}*\frac{2ab(a^2+b^2+a^2-b^2)}{(a^2-b^2)(a^2+b^2)}}= \frac{ (\frac{4a^2b^2}{a^4-b^4} )^2 }{\frac{2ab*2b^2}{a^4-b^4}*\frac{2ab*2a^2}{a^4-b^4}}=$ $= \frac{ \frac{16a^4b^4}{(a^4-b^4)^2} }{\frac{4ab^3}{a^4-b^4}*\frac{4a^3b}{a^4-b^4}}= \frac{16a^4b^4}{(a^4-b^4)^2} * \frac{(a^4-b^4)^2}{16a^4b^4} =1$