1. (★)计算:
(1) $(\dfrac{x^2}{-2y}) · \dfrac{6xy^2}{x^4} =$ ;
(2) $\dfrac{x^2 - 2x + 1}{x^2 - 1} ÷ \dfrac{x - 1}{x^2 + x} =$ .
(1) $(\dfrac{x^2}{-2y}) · \dfrac{6xy^2}{x^4} =$ ;
(2) $\dfrac{x^2 - 2x + 1}{x^2 - 1} ÷ \dfrac{x - 1}{x^2 + x} =$ .
答案
解:
(1) 原式$=\frac{x^2· 6xy^2}{-2y· x^4}$
$=\frac{6x^3y^2}{-2x^4y}$
$=-\frac{3y}{x}$
(2) 原式$=\frac{x^2-2x+1}{x^2-1}· \frac{x^2+x}{x-1}$
$=\frac{(x-1)^2}{(x+1)(x-1)}· \frac{x(x+1)}{x-1}$
$=x$
(1) 原式$=\frac{x^2· 6xy^2}{-2y· x^4}$
$=\frac{6x^3y^2}{-2x^4y}$
$=-\frac{3y}{x}$
(2) 原式$=\frac{x^2-2x+1}{x^2-1}· \frac{x^2+x}{x-1}$
$=\frac{(x-1)^2}{(x+1)(x-1)}· \frac{x(x+1)}{x-1}$
$=x$
2. (★)分式的乘方法则:分式乘方要把、分别乘方,即$(\dfrac{a}{b})^n =$($n$为正整数)
答案
解:
分子
分母
$\dfrac{a^n}{b^n}$
分子
分母
$\dfrac{a^n}{b^n}$
3. (★)由乘方的定义 $a^n = \underbrace{a · a · \dots · a}_{n个}$,得 $(-\dfrac{a}{b})^n = \underbrace{(-\dfrac{a}{b}) · (-\dfrac{a}{b}) · \dots · (-\dfrac{a}{b})}_{n个} = \underbrace{(-1) · (-1) · \dots · (-1)}_{n个} · \underbrace{\dfrac{a}{b} · \dfrac{a}{b} · \dots · \dfrac{a}{b}}_{n个} = \_\_\_\_\_\_.$
答案
解:
$\begin{aligned}&\underbrace{(-1) · (-1) · \dots · (-1)}_{n个} · \underbrace{\dfrac{a}{b} · \dfrac{a}{b} · \dots · \dfrac{a}{b}}_{n个} \\=& (-1)^n · (\frac{a}{b})^n \\=& \frac{(-1)^n a^n}{b^n}\end{aligned}$
最终答案为$\boldsymbol{\dfrac{(-1)^n a^n}{b^n}}$
$\begin{aligned}&\underbrace{(-1) · (-1) · \dots · (-1)}_{n个} · \underbrace{\dfrac{a}{b} · \dfrac{a}{b} · \dots · \dfrac{a}{b}}_{n个} \\=& (-1)^n · (\frac{a}{b})^n \\=& \frac{(-1)^n a^n}{b^n}\end{aligned}$
最终答案为$\boldsymbol{\dfrac{(-1)^n a^n}{b^n}}$
4. (★)计算:
(1)$(-2x^3y)^2 · (-3xy)^3 =$ ;
(2)$(-2x)^3 · (5x^2y)^2 =$ ;
(3)$(\dfrac{2}{a})^3 =$ ;
(4)$(\dfrac{4y}{3x})^2 =$ .
(1)$(-2x^3y)^2 · (-3xy)^3 =$ ;
(2)$(-2x)^3 · (5x^2y)^2 =$ ;
(3)$(\dfrac{2}{a})^3 =$ ;
(4)$(\dfrac{4y}{3x})^2 =$ .
答案
解:
(1) 先计算乘方:
$(-2x^3y)^2 = 4x^6y^2$,$(-3xy)^3 = -27x^3y^3$
原式$=4x^6y^2 · (-27x^3y^3) = -108x^9y^5$
(2) 先计算乘方:
$(-2x)^3 = -8x^3$,$(5x^2y)^2 = 25x^4y^2$
原式$=-8x^3 · 25x^4y^2 = -200x^7y^2$
(3) 根据分式乘方法则:
$(\dfrac{2}{a})^3 = \dfrac{2^3}{a^3} = \dfrac{8}{a^3}$
(4) 根据分式乘方法则:
$(\dfrac{4y}{3x})^2 = \dfrac{(4y)^2}{(3x)^2} = \dfrac{16y^2}{9x^2}$
最终答案依次为:$\boldsymbol{-108x^9y^5}$;$\boldsymbol{-200x^7y^2}$;$\boldsymbol{\dfrac{8}{a^3}}$;$\boldsymbol{\dfrac{16y^2}{9x^2}}$
(1) 先计算乘方:
$(-2x^3y)^2 = 4x^6y^2$,$(-3xy)^3 = -27x^3y^3$
原式$=4x^6y^2 · (-27x^3y^3) = -108x^9y^5$
(2) 先计算乘方:
$(-2x)^3 = -8x^3$,$(5x^2y)^2 = 25x^4y^2$
原式$=-8x^3 · 25x^4y^2 = -200x^7y^2$
(3) 根据分式乘方法则:
$(\dfrac{2}{a})^3 = \dfrac{2^3}{a^3} = \dfrac{8}{a^3}$
(4) 根据分式乘方法则:
$(\dfrac{4y}{3x})^2 = \dfrac{(4y)^2}{(3x)^2} = \dfrac{16y^2}{9x^2}$
最终答案依次为:$\boldsymbol{-108x^9y^5}$;$\boldsymbol{-200x^7y^2}$;$\boldsymbol{\dfrac{8}{a^3}}$;$\boldsymbol{\dfrac{16y^2}{9x^2}}$
5. (★★)计算:
(1) $\dfrac{ab^2}{6c^2} · \dfrac{-4c}{b^2} ÷ \dfrac{a}{c}$;
(2) $\dfrac{2}{3x-2} ÷ \dfrac{3}{4-9x^2} · \dfrac{x}{3x+2}$。
(1) $\dfrac{ab^2}{6c^2} · \dfrac{-4c}{b^2} ÷ \dfrac{a}{c}$;
(2) $\dfrac{2}{3x-2} ÷ \dfrac{3}{4-9x^2} · \dfrac{x}{3x+2}$。
答案
解:
(1) 原式$=\dfrac{ab^2}{6c^2} · \dfrac{-4c}{b^2} · \dfrac{c}{a}$
$=\dfrac{ab^2 · (-4c) · c}{6c^2 · b^2 · a}$
$=-\dfrac{4ab^2c^2}{6ab^2c^2}$
$=-\dfrac{2}{3}$
(2) 原式$=\dfrac{2}{3x-2} · \dfrac{4-9x^2}{3} · \dfrac{x}{3x+2}$
$=\dfrac{2}{3x-2} · \dfrac{(2-3x)(2+3x)}{3} · \dfrac{x}{3x+2}$
$=\dfrac{2}{3x-2} · \dfrac{-(3x-2)(3x+2)}{3} · \dfrac{x}{3x+2}$
$=-\dfrac{2x}{3}$
(1) 原式$=\dfrac{ab^2}{6c^2} · \dfrac{-4c}{b^2} · \dfrac{c}{a}$
$=\dfrac{ab^2 · (-4c) · c}{6c^2 · b^2 · a}$
$=-\dfrac{4ab^2c^2}{6ab^2c^2}$
$=-\dfrac{2}{3}$
(2) 原式$=\dfrac{2}{3x-2} · \dfrac{4-9x^2}{3} · \dfrac{x}{3x+2}$
$=\dfrac{2}{3x-2} · \dfrac{(2-3x)(2+3x)}{3} · \dfrac{x}{3x+2}$
$=\dfrac{2}{3x-2} · \dfrac{-(3x-2)(3x+2)}{3} · \dfrac{x}{3x+2}$
$=-\dfrac{2x}{3}$
6. (★★)计算:
(1) $(\dfrac{3y}{-2x})^2 · (\dfrac{2x}{3y})^3$;
(2) $\dfrac{x^2 - 4y^2}{x^2 + 2xy + y^2} ÷ (\dfrac{x + 2y}{x + y})^2$。
(1) $(\dfrac{3y}{-2x})^2 · (\dfrac{2x}{3y})^3$;
(2) $\dfrac{x^2 - 4y^2}{x^2 + 2xy + y^2} ÷ (\dfrac{x + 2y}{x + y})^2$。
答案
解:
(1)
$\begin{aligned}(\frac{3y}{-2x})^2 · (\frac{2x}{3y})^3&=\frac{(3y)^2}{(-2x)^2} · \frac{(2x)^3}{(3y)^3}\\&=\frac{9y^2}{4x^2} · \frac{8x^3}{27y^3}\\&=\frac{2x}{3y}\end{aligned}$
(2)
$\begin{aligned}\frac{x^2-4y^2}{x^2+2xy+y^2} ÷ (\frac{x+2y}{x+y})^2&=\frac{(x+2y)(x-2y)}{(x+y)^2} ÷ \frac{(x+2y)^2}{(x+y)^2}\\&=\frac{(x+2y)(x-2y)}{(x+y)^2} · \frac{(x+y)^2}{(x+2y)^2}\\&=\frac{x-2y}{x+2y}\end{aligned}$
(1)
$\begin{aligned}(\frac{3y}{-2x})^2 · (\frac{2x}{3y})^3&=\frac{(3y)^2}{(-2x)^2} · \frac{(2x)^3}{(3y)^3}\\&=\frac{9y^2}{4x^2} · \frac{8x^3}{27y^3}\\&=\frac{2x}{3y}\end{aligned}$
(2)
$\begin{aligned}\frac{x^2-4y^2}{x^2+2xy+y^2} ÷ (\frac{x+2y}{x+y})^2&=\frac{(x+2y)(x-2y)}{(x+y)^2} ÷ \frac{(x+2y)^2}{(x+y)^2}\\&=\frac{(x+2y)(x-2y)}{(x+y)^2} · \frac{(x+y)^2}{(x+2y)^2}\\&=\frac{x-2y}{x+2y}\end{aligned}$
7. (★★)计算:
(1)$\dfrac{2x^2}{3y^2} · \dfrac{5y}{6x} ÷ \dfrac{10x^2}{21y}$;
(2)$\dfrac{a+2}{a^2-2a+1} · \dfrac{a^2-4a+4}{a+1} ÷ \dfrac{a^2-4}{a^2-1}$
(1)$\dfrac{2x^2}{3y^2} · \dfrac{5y}{6x} ÷ \dfrac{10x^2}{21y}$;
(2)$\dfrac{a+2}{a^2-2a+1} · \dfrac{a^2-4a+4}{a+1} ÷ \dfrac{a^2-4}{a^2-1}$
答案
解:
(1) 原式$=\dfrac{2x^2}{3y^2} · \dfrac{5y}{6x} · \dfrac{21y}{10x^2}$
$=\dfrac{2x^2 · 5y · 21y}{3y^2 · 6x · 10x^2}$
$=\dfrac{210x^2y^2}{180x^3y^2}$
$=\dfrac{7}{6x}$
(2) 先对各多项式因式分解:
$a^2-2a+1=(a-1)^2$,$a^2-4a+4=(a-2)^2$,$a^2-4=(a+2)(a-2)$,$a^2-1=(a+1)(a-1)$
原式$=\dfrac{a+2}{(a-1)^2} · \dfrac{(a-2)^2}{a+1} · \dfrac{(a+1)(a-1)}{(a+2)(a-2)}$
约去分子分母的公因式:
$=\dfrac{a-2}{a-1}$
(1) 原式$=\dfrac{2x^2}{3y^2} · \dfrac{5y}{6x} · \dfrac{21y}{10x^2}$
$=\dfrac{2x^2 · 5y · 21y}{3y^2 · 6x · 10x^2}$
$=\dfrac{210x^2y^2}{180x^3y^2}$
$=\dfrac{7}{6x}$
(2) 先对各多项式因式分解:
$a^2-2a+1=(a-1)^2$,$a^2-4a+4=(a-2)^2$,$a^2-4=(a+2)(a-2)$,$a^2-1=(a+1)(a-1)$
原式$=\dfrac{a+2}{(a-1)^2} · \dfrac{(a-2)^2}{a+1} · \dfrac{(a+1)(a-1)}{(a+2)(a-2)}$
约去分子分母的公因式:
$=\dfrac{a-2}{a-1}$
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