Question

Выполните действия пж​

Answer 1

1)

$(a^2-1)*(\frac{1}{a-1}-\frac{1}{a+1}+1)=a^2+1\\\\1)\ \frac{1}{a-1}-\frac{1}{a+1}+1=\frac{a+1-a+1+a^2-1}{(a-1)(a+1)}=\frac{a^2+1}{a^2-1}\\\\2)\ \frac{(a^2-1)*(a^2+1)}{1*(a^2-1)}=a^2+1$

2)

$(\frac{1}{y}+\frac{2}{x-y})*(x-\frac{x^2+y^2}{x+y})=1\\\\1)\ \frac{1}{y}+\frac{2}{x-y}=\frac{x-y+2y}{y(x-y)}=\frac{x+y}{y(x-y)}\\\\2)\ x-\frac{x^2+y^2}{x+y}=\frac{x(x+y)-x^2-y^2}{x+y}=\frac{y(x-y)}{x+y}\\\\3)\ \frac{(x+y)*y(x-y)}{y(x-y)*(x+y)}=1$

3)

$(x+1-\frac{1}{1-x}):(x-\frac{x^2}{x-1})=-x\\\\1)\ x+1-\frac{1}{1-x}=\frac{x(1-x)+1-x-1}{1-x}=\frac{-x^2}{1-x}\\\\2)\ x-\frac{x^2}{x-1}=\frac{x(x-1)-x^2}{x-1}=\frac{-x}{x-1}\\\\3)\ \frac{-x^2}{1-x}:\frac{-x}{x-1}=-\frac{-x^2}{x-1}:\frac{-x}{x-1}=-\frac{-x^2*(x-1)}{(x-1)*(-x)}=-x$

4)

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

Ответ: 1) a^2+1; 2) 1; 3) -x; 4) a.