例2 计算:
(1) $\frac{5}{6ab} - \frac{2}{3ac} + \frac{3}{4abc}$;
(2) $\frac{12}{m^2 - 9} + \frac{2}{3 - m}$。
解:(1) $\frac{5}{6ab} - \frac{2}{3ac} + \frac{3}{4abc} = \frac{10c}{12abc} - \frac{8b}{12abc} + \frac{9}{12abc} = \frac{10c - 8b + 9}{12abc}$;
(2) $\frac{12}{m^2 - 9} + \frac{2}{3 - m} = \frac{12}{(m + 3)(m - 3)} - \frac{2}{m - 3}$
将分母先分解因式,便于确定最简公分母。
$= \frac{12}{(m + 3)(m - 3)} - \frac{2(m + 3)}{(m - 3)(m + 3)}$
$= \frac{12 - 2(m + 3)}{(m + 3)(m - 3)} = \frac{-2m + 6}{(m + 3)(m - 3)}$
$= \frac{-2(m - 3)}{(m + 3)(m - 3)}$
当分子能分解因式时,要分解因式,便于确定是否能继续约分。
$= -\frac{2}{m + 3}$
结果要化成最简分式或整式。
当分母不同时,先通分,再加减。
(1) $\frac{5}{6ab} - \frac{2}{3ac} + \frac{3}{4abc}$;
(2) $\frac{12}{m^2 - 9} + \frac{2}{3 - m}$。
解:(1) $\frac{5}{6ab} - \frac{2}{3ac} + \frac{3}{4abc} = \frac{10c}{12abc} - \frac{8b}{12abc} + \frac{9}{12abc} = \frac{10c - 8b + 9}{12abc}$;
(2) $\frac{12}{m^2 - 9} + \frac{2}{3 - m} = \frac{12}{(m + 3)(m - 3)} - \frac{2}{m - 3}$
将分母先分解因式,便于确定最简公分母。
$= \frac{12}{(m + 3)(m - 3)} - \frac{2(m + 3)}{(m - 3)(m + 3)}$
$= \frac{12 - 2(m + 3)}{(m + 3)(m - 3)} = \frac{-2m + 6}{(m + 3)(m - 3)}$
$= \frac{-2(m - 3)}{(m + 3)(m - 3)}$
当分子能分解因式时,要分解因式,便于确定是否能继续约分。
$= -\frac{2}{m + 3}$
结果要化成最简分式或整式。
当分母不同时,先通分,再加减。
答案:
例3 计算:
(1)$(\dfrac{2a}{b})^2 · \dfrac{1}{a - b} - \dfrac{a}{b} ÷ \dfrac{b}{4}$;
(2)$(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}) ÷ \dfrac{x - 4}{x}$。
解:(1)$(\dfrac{2a}{b})^2 · \dfrac{1}{a - b} - \dfrac{a}{b} ÷ \dfrac{b}{4} = \dfrac{4a^2}{b^2} · \dfrac{1}{a - b} - \dfrac{a}{b} · \dfrac{4}{b}$
$\xleftarrow{\mathrm{先乘方,再乘除.}} \quad = \dfrac{4a^2}{b^2(a - b)} - \dfrac{4a}{b^2} = \dfrac{4a^2}{b^2(a - b)} - \dfrac{4a(a - b)}{b^2(a - b)}$
$\xleftarrow{\mathrm{先通分,再相减.}} \quad = \dfrac{4a^2 - 4a^2 + 4ab}{b^2(a - b)}$
$\xleftarrow{\mathrm{合并同类项.}} \quad = \dfrac{4ab}{b^2(a - b)}$
$\xleftarrow{\mathrm{约分,结果要化成最简分式或整式.}} \quad = \dfrac{4a}{ab - b^2}$;
(2)$(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}) ÷ \dfrac{x - 4}{x} = [\dfrac{x + 2}{x(x - 2)} - \dfrac{x - 1}{(x - 2)^2}] · \dfrac{x}{x - 4}$
$\xleftarrow{\mathrm{除式与括号里的分式有联系时,可先化除为乘,使计算简便.}} \quad = \dfrac{(x + 2)(x - 2) - (x - 1)x}{x(x - 2)^2} · \dfrac{x}{x - 4}$
$\quad = \dfrac{x^2 - 4 - x^2 + x}{(x - 2)^2(x - 4)}$
$\xleftarrow{\mathrm{合并同类项后再约分,结果要化成最简分式或整式.}} \quad = \dfrac{1}{(x - 2)^2}$。
(1)$(\dfrac{2a}{b})^2 · \dfrac{1}{a - b} - \dfrac{a}{b} ÷ \dfrac{b}{4}$;
(2)$(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}) ÷ \dfrac{x - 4}{x}$。
解:(1)$(\dfrac{2a}{b})^2 · \dfrac{1}{a - b} - \dfrac{a}{b} ÷ \dfrac{b}{4} = \dfrac{4a^2}{b^2} · \dfrac{1}{a - b} - \dfrac{a}{b} · \dfrac{4}{b}$
$\xleftarrow{\mathrm{先乘方,再乘除.}} \quad = \dfrac{4a^2}{b^2(a - b)} - \dfrac{4a}{b^2} = \dfrac{4a^2}{b^2(a - b)} - \dfrac{4a(a - b)}{b^2(a - b)}$
$\xleftarrow{\mathrm{先通分,再相减.}} \quad = \dfrac{4a^2 - 4a^2 + 4ab}{b^2(a - b)}$
$\xleftarrow{\mathrm{合并同类项.}} \quad = \dfrac{4ab}{b^2(a - b)}$
$\xleftarrow{\mathrm{约分,结果要化成最简分式或整式.}} \quad = \dfrac{4a}{ab - b^2}$;
(2)$(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}) ÷ \dfrac{x - 4}{x} = [\dfrac{x + 2}{x(x - 2)} - \dfrac{x - 1}{(x - 2)^2}] · \dfrac{x}{x - 4}$
$\xleftarrow{\mathrm{除式与括号里的分式有联系时,可先化除为乘,使计算简便.}} \quad = \dfrac{(x + 2)(x - 2) - (x - 1)x}{x(x - 2)^2} · \dfrac{x}{x - 4}$
$\quad = \dfrac{x^2 - 4 - x^2 + x}{(x - 2)^2(x - 4)}$
$\xleftarrow{\mathrm{合并同类项后再约分,结果要化成最简分式或整式.}} \quad = \dfrac{1}{(x - 2)^2}$。
答案: