《2025年阳光课堂金牌练习册八年级数学上册人教版》

【典型例题1】计算：
(1)$\dfrac{-a^{2}b}{2c}\cdot\left(-\dfrac{4cd}{5ab^{2}}\right)$；(2)$\dfrac{ab^{3}}{2c^{2}}÷\dfrac{-5a^{2}b^{2}}{4cd}$；
(3)$\dfrac{x^{2}-4x + 4}{x^{2}+2x + 1}\cdot\dfrac{x^{2}+x}{x^{2}-4}$；
(4)$\dfrac{-x^{2}+2x - 1}{2 + x}÷\dfrac{x^{2}-1}{x^{2}+2x}$.

答案： 【解】
(1)$\dfrac{-a^{2}b}{2c}\cdot\left(-\dfrac{4cd}{5ab^{2}}\right)= \dfrac{a^{2}b}{2c}\cdot\dfrac{4cd}{5ab^{2}}= \dfrac{a^{2}b\cdot4cd}{2c\cdot5ab^{2}}= \dfrac{2ad}{5b}$.
(2)$\dfrac{ab^{3}}{2c^{2}}÷\dfrac{-5a^{2}b^{2}}{4cd}= \dfrac{ab^{3}}{2c^{2}}\cdot\dfrac{4cd}{-5a^{2}b^{2}}= -\dfrac{ab^{3}\cdot4cd}{2c^{2}\cdot5a^{2}b^{2}}= -\dfrac{2bd}{5ac}$.
(3)$\dfrac{x^{2}-4x + 4}{x^{2}+2x + 1}\cdot\dfrac{x^{2}+x}{x^{2}-4}= \dfrac{(x - 2)^{2}}{(x + 1)^{2}}\cdot\dfrac{x(x + 1)}{(x + 2)(x - 2)}= \dfrac{x(x - 2)^{2}(x + 1)}{(x + 1)^{2}(x + 2)(x - 2)}= \dfrac{x(x - 2)}{(x + 1)(x + 2)}$.
(4)$\dfrac{-x^{2}+2x - 1}{2 + x}÷\dfrac{x^{2}-1}{x^{2}+2x}= \dfrac{-(x - 1)^{2}}{x + 2}\cdot\dfrac{x(x + 2)}{(x + 1)(x - 1)}= -\dfrac{x(x - 1)}{x + 1}$.

1. 计算：
(1)$\dfrac{4a + 4b}{5ab}\cdot\dfrac{15a^{2}b}{a^{2}-b^{2}}$；
(2)$\dfrac{5m^{2}}{6n}÷(-5mn^{2})$；
(3)$\dfrac{a^{2}-4}{a - 3}÷\dfrac{a + 2}{a^{2}-6a + 9}$.

答案：
(1)
原式 $=\dfrac{4(a + b)}{5ab}\cdot\dfrac{15a^{2}b}{(a + b)(a - b)}$
$=\dfrac{4(a + b)×15a^{2}b}{5ab×(a + b)(a - b)}$
$=\dfrac{12a}{a - b}$
(2)
原式 $=\dfrac{5m^{2}}{6n}×\dfrac{1}{-5mn^{2}}$
$=\dfrac{5m^{2}}{6n×(-5mn^{2})}$
$=-\dfrac{m}{6n^{3}}$
(3)
原式 $=\dfrac{a^{2}-4}{a - 3}×\dfrac{a^{2}-6a + 9}{a + 2}$
$=\dfrac{(a + 2)(a - 2)}{a - 3}×\dfrac{(a - 3)^{2}}{a + 2}$
$=(a - 2)(a - 3)$
$=a^{2}-5a + 6$