Пожалуйста подробно решите примеры.

[latex] \frac{1+x}{1-x}+ \frac{1-x}{1+x}- \frac{2x^2}{1-x^2}= \frac{1+x}{1-x}*1+ \frac{1-x}{1+x}*1- \frac{2x^2}{1^2-x^2}=[/latex] [latex]= \frac{1+x}{1-x}* \frac{1+x}{1+x} + \frac{1-x}{1+x}* \frac{1-x}{1-x} - \frac{2x^2}{(1-x)(1+x)}=[/latex] [latex]= \frac{(1+x)*(1+x)}{(1-x)*(1+x)} + \frac{(1-x)*(1-x)}{(1+x)*(1-x)} - \frac{2x^2}{(1-x)(1+x)}=[/latex] [latex]= \frac{(1+x)^2}{(1-x)(1+x)} + \frac{(1-x)^2}{(1-x)(1+x)} - \frac{2x^2}{(1-x)(1+x)}=[/latex] [latex]= \frac{(1+x)^2+(1-x)^2-2x^2}{(1-x)(1+x)} = \frac{1^2+2*1*x+x^2+(1^2-2*1*x+x^2)-2x^2}{(1-x)(1+x)}=[/latex] [latex]= \frac{1+2x+x^2+1-2x+x^2-2x^2}{(1-x)(1+x)} = \frac{x^2+x^2-2x^2+2x-2x+1+1}{(1-x)(1+x)} = \frac{0*x^2+0*x+2}{(1-x)(1+x)}=[/latex] [latex]= \frac{2}{(1-x)(1+x)}[/latex] --------------------------------------- [latex]\frac{1}{a+b}-\frac{2b}{a^2-b^2}+\frac{2}{a-b}=[/latex] [latex]=\frac{1*(a-b)}{(a+b)*(a-b)}-\frac{2b}{(a-b)(a+b)}+\frac{2*(a+b)}{(a-b)*(a+b)}=[/latex] [latex]=\frac{a-b}{(a+b)(a-b)}-\frac{2b}{(a+b)(a-b)}+\frac{2(a+b)}{(a+b)(a-b)}=[/latex] [latex]=\frac{a-b-2b+2(a+b)}{(a+b)(a-b)}=[/latex] [latex]=\frac{a-b-2b+2a+2b}{(a+b)(a-b)}=[/latex] [latex]=\frac{3a-b}{(a+b)(a-b)}[/latex]