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1. 下列各式是最简分式的是(
A.$\frac{4y+2x}{4a}$
B.$\frac{y-x}{x-y}$
C.$\frac{x^2+1}{x-1}$
D.$\frac{x^2-1}{x+1}$
C
).A.$\frac{4y+2x}{4a}$
B.$\frac{y-x}{x-y}$
C.$\frac{x^2+1}{x-1}$
D.$\frac{x^2-1}{x+1}$
答案
1.C
2. 将分式$\frac{3x^2}{x^2 - y^2}$与$\frac{x}{2(x - y)}$通分后，分式$\frac{3x^2}{x^2 - y^2}$的分子应变为(
A.$6x^2(x - y)^2$
B.$2(x - y)$
C.$6x^2$
D.$6x^2(x + y)$
C
).A.$6x^2(x - y)^2$
B.$2(x - y)$
C.$6x^2$
D.$6x^2(x + y)$
答案
2.C
3. 对分式$\frac{1}{2a^2b}$和$\frac{1}{3ab^3}$进行通分，它们的最简公分母是
$6a^2b^3$
。答案
3. $6a^2b^3$
4. 化简下列分式：
(1)$\dfrac{12x^2y^3}{9x^3y^2}$；(2)$\dfrac{x-y}{(y-x)^3}$.
(1)$\dfrac{12x^2y^3}{9x^3y^2}$；(2)$\dfrac{x-y}{(y-x)^3}$.
答案
解 (1)$\frac{12x^2 y^3}{9x^3 y^2} = \frac{4y · 3x^2 y^2}{3x · 3x^2 y^2} = \frac{4y}{3x}.$
(2)$\frac{x-y}{(y-x)^3} = \frac{x-y}{-(x-y)^3}$
$= \frac{(x-y)}{-(x-y)^2 · (x-y)}$
$= -\frac{1}{(x-y)^2}.$
(2)$\frac{x-y}{(y-x)^3} = \frac{x-y}{-(x-y)^3}$
$= \frac{(x-y)}{-(x-y)^2 · (x-y)}$
$= -\frac{1}{(x-y)^2}.$
5. 通分：
(1) $\frac{a}{2b}, \frac{2}{5a^2b^2c}$；
(2) $\frac{1}{x^2 - x}, \frac{-1}{x^2 - 2x + 1}$。
(1) $\frac{a}{2b}, \frac{2}{5a^2b^2c}$；
(2) $\frac{1}{x^2 - x}, \frac{-1}{x^2 - 2x + 1}$。
答案
解 (1)最简公分母是$10a^2b^2c$.
$\frac{a}{2b} = \frac{5a^3 bc}{10a^2b^2c},$
$\frac{2}{5a^2b^2c} = \frac{4}{10a^2b^2c}.$
(2)最简公分母是$x(x-1)^2$.
$\frac{1}{x^2 - x} = \frac{x-1}{x(x-1)^2},$
$\frac{-1}{x^2 - 2x + 1} = -\frac{x}{x(x-1)^2}.$
$\frac{a}{2b} = \frac{5a^3 bc}{10a^2b^2c},$
$\frac{2}{5a^2b^2c} = \frac{4}{10a^2b^2c}.$
(2)最简公分母是$x(x-1)^2$.
$\frac{1}{x^2 - x} = \frac{x-1}{x(x-1)^2},$
$\frac{-1}{x^2 - 2x + 1} = -\frac{x}{x(x-1)^2}.$
6.先化简,再求值:$\dfrac{1-4x^2}{2x^2+x}$,其中$x=-1$.
答案
解 $\frac{1-4x^2}{2x^2+x} = \frac{1^2-(2x)^2}{x(2x+1)}$
$= \frac{(1+2x)(1-2x)}{x(2x+1)}$
$= \frac{1-2x}{x}.$
当$x=-1$时，
原式$= \frac{1-2×(-1)}{-1} = -3.$
$= \frac{(1+2x)(1-2x)}{x(2x+1)}$
$= \frac{1-2x}{x}.$
当$x=-1$时，
原式$= \frac{1-2×(-1)}{-1} = -3.$
7.定义:若一个分式约分后是一个整式,则称这个分式为“巧分式”,约分后的整式称为这个分式的“巧整式”.例如:
$\frac{4x^2 -8x}{x-2} = \frac{4x(x-2)}{x-2}=4x$,则称分式$\frac{4x^2 -8x}{x-2}$是“巧分式”,$4x$为它的“巧整式”.根据上述定义,解决下列问题.
(1)若分式$\frac{x^2 -4x +m}{x+n}$($m,n$为常数)是一个“巧分式”,它的“巧整式”为$x-7$,求$m,n$的值;
(2)若分式$\frac{-2x^3 +2x}{A}$的“巧整式”为$1-x$,请判断$\frac{2x^3 +4x^2 +2x}{A}$是不是“巧分式”,并说明理由.
$\frac{4x^2 -8x}{x-2} = \frac{4x(x-2)}{x-2}=4x$,则称分式$\frac{4x^2 -8x}{x-2}$是“巧分式”,$4x$为它的“巧整式”.根据上述定义,解决下列问题.
(1)若分式$\frac{x^2 -4x +m}{x+n}$($m,n$为常数)是一个“巧分式”,它的“巧整式”为$x-7$,求$m,n$的值;
(2)若分式$\frac{-2x^3 +2x}{A}$的“巧整式”为$1-x$,请判断$\frac{2x^3 +4x^2 +2x}{A}$是不是“巧分式”,并说明理由.
答案
解 (1)$\because$ 分式$\frac{x^2 -4x +m}{x+n}$($m,n$为常数)是一个“巧分式”,它的“巧整式”为$x-7$,
$\therefore (x+n)(x-7) = x^2 -4x +m$,
$\therefore x^2 + (n-7)x -7n = x^2 -4x +m$,
$\therefore n-7=-4, m=-7n$,
$\therefore n=3, m=-21.$
(2)$\frac{2x^3 +4x^2 +2x}{A}$是“巧分式”.
理由如下：
$\because$ 分式$\frac{-2x^3 +2x}{A}$的“巧整式”为$1-x$,
$\therefore A = \frac{-2x^3 +2x}{1-x} = \frac{2x(1-x^2)}{1-x} = \frac{2x(1-x)(1+x)}{1-x} = 2x(1+x)$,即$A=2x^2+2x$,
则$\frac{2x^3 +4x^2 +2x}{A} = \frac{2x^3 +4x^2 +2x}{2x^2 +2x}$
$= \frac{2x(x^2 +2x +1)}{2x(x+1)} = \frac{(x+1)^2}{(x+1)}$
$= x+1$,
又$x+1$是整式，
$\therefore \frac{2x^3 +4x^2 +2x}{A}$是“巧分式”.
$\therefore (x+n)(x-7) = x^2 -4x +m$,
$\therefore x^2 + (n-7)x -7n = x^2 -4x +m$,
$\therefore n-7=-4, m=-7n$,
$\therefore n=3, m=-21.$
(2)$\frac{2x^3 +4x^2 +2x}{A}$是“巧分式”.
理由如下：
$\because$ 分式$\frac{-2x^3 +2x}{A}$的“巧整式”为$1-x$,
$\therefore A = \frac{-2x^3 +2x}{1-x} = \frac{2x(1-x^2)}{1-x} = \frac{2x(1-x)(1+x)}{1-x} = 2x(1+x)$,即$A=2x^2+2x$,
则$\frac{2x^3 +4x^2 +2x}{A} = \frac{2x^3 +4x^2 +2x}{2x^2 +2x}$
$= \frac{2x(x^2 +2x +1)}{2x(x+1)} = \frac{(x+1)^2}{(x+1)}$
$= x+1$,
又$x+1$是整式，
$\therefore \frac{2x^3 +4x^2 +2x}{A}$是“巧分式”.
