$解：(1)\frac {1}{x+1}=\frac {(x+1)(x-1)}{(x+1)²(x-1)}；\frac {x-1}{x²+2x+1}=\frac {(x-1)²}{(x+1)²(x-1)}$
$\frac {1}{x-1}=\frac {(x+1)²}{(x+1)²(x-1)}$
$(2)\frac {1}{x-1}=\frac {x(x+1)}{x(x+1)(x-1)}；\frac {1}{x²-1}=\frac {x}{x(x+1)(x-1)}$
$\frac {1}{x²+x}=\frac {x-1}{x(x+1)(x-1)}$
