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第4页

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解：$\frac 1{ax}=\frac {b}{abx}，$$\frac 1{bx}=\frac {a}{abx}$

解：$\frac 3{2x^2y}=\frac {9y}{6x^2y^2}$

$\frac {5}{3xy^2}=\frac {10x}{6x^2y^2}$

解：$\frac {x-1}{3x^2}=\frac {a(x-1)}{3ax^2}，$

$\frac 2{ax}=\frac {6x}{3ax^2}$

解：$\frac {a}{2b}=\frac {6a^2}{12ab}，$

$\frac b{3a}=\frac {4b^2}{12ab}，$$\frac {c}{4ab}=\frac {3c}{12ab}$

解：$\frac {4a}{5b^2c}=\frac {8a^3c}{10a^2b^2c^2}，$

$\frac {3c}{10a^2b}=\frac {3bc^3}{10a^2b^2c^2}，$

$\frac {5b}{-2ac^2}=-\frac {25ab^3}{10a^2b^2c^2}$

解：$\frac {3a}{2a-b}，$

$-\frac 1{b-2a}=\frac 1{2a-b}$

解：$\frac 2{x+1}=\frac {2(x+2)}{(x+1)(x+2)}，$

$\frac 3{x+2}=\frac {3(x+1)}{(x+1)(x+2)}$

解：$\frac {b}{a-x}=\frac {by}{ay-xy}，$$\frac {c}{ay-xy}$

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

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

解：$\frac 2{9-3a}=-\frac {2(a+3)}{3(a-3)(a+3)}，$

$\frac {a-1}{a^2-9}=\frac {3(a-1)}{3(a-3)(a+3)}$

解：$\frac {2a}{a^2-9}=\frac {2a(a-3)}{(a-3)^2(a+3)}，$

$\frac 3{a^2-6a+9}=\frac {3(a+3)}{(a-3)^2(a+3)}$

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

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

解：$\frac 1{x^2-4}=\frac 1{(x+2)(x-2)}=\frac 2{2(x+2)(x-2)}，$

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

解：$\frac 1{x-1}=\frac {x(x+1)}{x(x+1)(x-1)}，$

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