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第92页

信息发布者：

$a-1$

$x-1$

解：原式$=\frac {3x+4-1}{x+1}$

$=\frac {3x+3}{x+1}$

$=\frac {3(x+1)}{x+1}$

$=3$

解：原式$=\frac {2x}{x^2-y^2}-\frac {5x+3y}{x^2-y^2}$

$=\frac {2x-(5x+3y)}{x^2-y^2}$

$=\frac {2x-5x-3y}{(x+y)(x-y)}$

$=\frac {-3(x+y)}{(x+y)(x-y)}$

$=-\frac {3}{x-y}$

解：原式$=\frac {m-6}{(m+2)(m-2)}+\frac {1}{m-2}$

$=\frac {m-6}{(m+2)(m-2)}+\frac {m+2}{(m+2)(m-2)}$

$=\frac {m-6+m+2}{(m+2)(m-2)}$

$=\frac {2m-4}{(m+2)(m-2)}$

$=\frac {2(m-2)}{(m+2)(m-2)}$

$=\frac {2}{m+2}$

解：原式$=\frac {(a+b)(a-b)}{ab}-\frac {b(a-b)}{-a(a-b)}$

$=\frac {(a+b)(a-b)}{ab}+\frac {b}{a}$

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

$=\frac {a^2-b^2+b^2}{ab}$

$=\frac {a^2}{ab}$

$=\frac {a}{b}$

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

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

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

$ $当$a=2$时，原式$=\frac {2+1}{2-1}=3$

解：原式$=\frac {2n}{m+2n}+\frac {m}{2n-m}+\frac {4mn}{(2n+m)(2n-m)}$

$=\frac {2n(2n-m)}{(2n+m)(2n-m)}+\frac {m(2n+m)}{(2n+m)(2n-m)}+\frac {4mn}{(2n+m)(2n-m)}$

$=\frac {4n^2-2mn+2mn+\mathrm {m^2}+4mn}{(2n+m)(2n-m)}$

$ =\frac {4n^2+4mn+\mathrm {m^2}}{(2n+m)(2n-m)}$

$ =\frac {(2n+m)^2}{(2n+m)(2n-m)}$

$ =\frac {2n+m}{2n-m}$

∵$\frac {m}{n}=\frac {1}{5}，$

∴$n=5m$

$ $原式$=\frac {2×5m+m}{2×5m-m}=\frac {10m+m}{10m-m}=\frac {11m}{9m}=\frac {11}{9}$