Question

Помогите с самостоятельной по алгебре

Answer 1

$1)\; \; (\frac{x+2}{x^2-1}+1):(1-\frac{x}{x+1})=\frac{x+1+x^2-1}{(x-1)(x+1)}:\frac{x+1-x}{x+1}=\\\\=\frac{x(x+1)}{(x-1)(x+1)}\cdot \frac{x+1}{1}=\frac{x(x+1)}{x-1}$

$2)\; \; (2n^2-2)\cdot (\frac{1}{n-1}-\frac{1}{n+1}-1)=2(n^2-1)\cdot \frac{n+1-(n-1)-(n^2-1)}{(n-1)(n+1)}=\\\\=2(n-1)(n+1)\cdot \frac{n+1-n+1-n^2+1}{(n-1)(n+1)}=2\cdot (3-n^2)$

$3)\; \; y=\frac{1}{3}\\\\(\frac{y-4}{3y-3}+\frac{1}{y-1}):\frac{y+1}{3}+\frac{2}{y^2-1}=\frac{y-4+3}{3(y-1)}:\frac{y+1}{3}+\frac{2}{(y-1)(y+1)}=\\\\=\frac{y-1}{3(y-1)}\cdot \frac{3}{y+1}+\frac{2}{(y-1)(y+1)}=\frac{1}{y+1}+\frac{2}{(y-1)(y+1)}=\frac{y-1+2}{(y-1)(y+1)}=\\\\=\frac{y+1}{(y-1)(y+1)}=\frac{1}{y-1}=\frac{1}{\frac{1}{3}-1}=\frac{3}{1-3}=-\frac{3}{2}=-1,5$