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第112页

信息发布者：

$ A$

$ C$

$ab$

$a+1$

-4

解：$(1)$原式$=\frac {1}{m+1}·\frac {(m+1)^2}{\mathrm {m^3}}-\frac {1}{\mathrm {m^3}}$

$=\frac {m+1}{\mathrm {m^3}}-\frac {1}{\mathrm {m^3}}$

$=\frac {m}{\mathrm {m^3}}$

$=\frac {1}{\mathrm {m^2}}$

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

$=\frac {(x+1)(x-1)}{x}÷\frac {x^2+2x+1}{x}$

$ =\frac {(x+1)(x-1)}{x}÷\frac {(x+1)^2}{x}$

$=\frac {(x+1)(x-1)}{x}·\frac {x}{(x+1)^2}$

$=\frac {x-1}{x+1}$

解：$(3)$原式$=\frac {y^2+x^2-2xy}{x}·\frac {x}{x^2-y^2}$

$=\frac {(x-y)^2}{x}·\frac {x}{(x+y)(x-y)}$

$=\frac {x-y}{x+y}$

解：$(4)$原式$=[\frac {(a-1)(a+2)}{a+2}+\frac {a+3}{a+2}]·\frac {a+2}{(a+1)(a-1)}$

$ =\frac {a^2-a+2a-2+a+3}{a+2}·\frac {a+2}{(a+1)(a-1)}$

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

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

$=\frac {a+1}{a-1}$

解：$(1)$原式$=\frac {a^2+2a+1}{a}÷\frac {a^2-1}{a}$

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

$=\frac {a+1}{a-1}$，

$ $当$a=2$时，原式$=\frac {2+1}{2-1}=3$

解：$(2)$原式$=\frac {a}{a-b}÷\frac {(a+b)(a-b)}{(a-b)^2}-\frac {a-b}{a+b}$

$=\frac {a}{a-b}·\frac {(a-b)^2}{(a+b)(a-b)}-\frac {a-b}{a+b}$

$ =\frac {a}{a+b}-\frac {a-b}{a+b}$

$=\frac {b}{a+b}$，

∵$b-2a=0$，

∴$b=2a$，

∴原式$=\frac {2a}{a+2a}=\frac {2}{3}$