Question

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Answer 1

$\displaystyle \tt 1). \ \ \frac{y^{2}-2y+1}{x-2}:\frac{y^{2}-1}{x^{2}-4}-\frac{2y-x}{y+1}=\frac{(y-1)^{2}(x-2)(x+2)}{(x-2)(y-1)(y+1)}-\frac{2y-x}{y+1}=\\\\\\{} \ =\frac{(y-1)(x+2)-2y+x}{y+1}=\frac{xy-x+2y-2-2y+x}{y+1}=\frac{xy-2}{y+1};\\\\\\2). \ \ (2n^{2}-2)\cdot\bigg(\frac{1}{n-1}-\frac{1}{n+1}-1\bigg)=\\\\\\{} \ =2\cdot(n-1)(n+1)\cdot\frac{(n+1)-(n-1)-(n-1)(n+1)}{(n-1)(n+1)}=\\\\\\{} \ =2\cdot(2-n^{2}+1)=-2n^{2}+6;$  

$\displaystyle \tt 3). \ \ \bigg(\frac{2x+2}{x^{2}-1}+1\bigg)\cdot\bigg(1-\frac{x}{x+1}\bigg)=\\\\\\{} \ \ =\frac{2\cdot(x+1)+(x-1)(x+1)}{(x-1)(x+1)}\cdot\frac{1}{x+1}=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1};$