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8. 式子$\frac{a - b}{(b - c)(c - a)}+\frac{b - c}{(a - b)(c - a)}+\frac{c - a}{(a - b)(b - c)}$的值不可能等于(
A.$-2$
B.$-1$
C.$0$
D.$1$
C
)A.$-2$
B.$-1$
C.$0$
D.$1$
答案
8. C
9. 在计算$\frac{2}{1 + 2x + x^{2}}+\frac{1}{2x + 2}$通分时，分母确定为
$ 2(x + 1)^{2} $
.答案
9. $ 2(x + 1)^{2} $
10. (2025·宿城期末)若$\frac{3}{n}-\frac{1}{m}=2$，则分式$\frac{-6m - 5mn + 2n}{3m - n}$的值为
$ -\frac{9}{2} $
.答案
10. $ -\frac{9}{2} $
11. 把下列各式通分：
(1) $\frac{1}{x^{2}-x},\frac{2}{x^{2}-1}$；
(2) $\frac{1}{a^{2}b - 4b},\frac{1}{b(a - 2)^{2}}$；
(3) $\frac{y}{x(x - y)^{2}},\frac{x}{y(y - x)^{2}}$；
(4) $\frac{5}{4 - 9x^{2}},\frac{2m}{4x^{2}-12x + 9}$；
(5) $\frac{x}{x - y},\frac{x}{x^{2}+2xy + y^{2}},\frac{2}{y^{2}-x^{2}}$；
(6) $\frac{2}{9 - 3a},\frac{1}{a^{2}-6a + 9},\frac{2}{3a^{2}-27}$.
(1) $\frac{1}{x^{2}-x},\frac{2}{x^{2}-1}$；
(2) $\frac{1}{a^{2}b - 4b},\frac{1}{b(a - 2)^{2}}$；
(3) $\frac{y}{x(x - y)^{2}},\frac{x}{y(y - x)^{2}}$；
(4) $\frac{5}{4 - 9x^{2}},\frac{2m}{4x^{2}-12x + 9}$；
(5) $\frac{x}{x - y},\frac{x}{x^{2}+2xy + y^{2}},\frac{2}{y^{2}-x^{2}}$；
(6) $\frac{2}{9 - 3a},\frac{1}{a^{2}-6a + 9},\frac{2}{3a^{2}-27}$.
答案
11. 解：(1) $ \frac{x + 1}{x(x - 1)(x + 1)},\frac{2x}{x(x - 1)(x + 1)} $．
(2) $ \frac{a - 2}{b(a - 2)^{2}(a + 2)},\frac{a + 2}{b(a - 2)^{2}(a + 2)} $．
(3) $ \frac{y^{2}}{xy(x - y)^{2}},\frac{x^{2}}{xy(x - y)^{2}} $．
(4) $ \frac{5(2x - 3)^{2}}{(2 - 3x)(2 + 3x)(2x - 3)^{2}},\frac{2m(2 - 3x)(2 + 3x)}{(2 - 3x)(2 + 3x)(2x - 3)^{2}} $．
(5) $ \frac{x(x + y)^{2}}{(x - y)(x + y)^{2}},\frac{x(x - y)}{(x - y)(x + y)^{2}} $，
$ \frac{-2(x + y)}{(x - y)(x + y)^{2}} $．
(6) $ \frac{-2(a - 3)(a + 3)}{3(a - 3)^{2}(a + 3)},\frac{3(a + 3)}{3(a - 3)^{2}(a + 3)} $，
$ \frac{2(a - 3)}{3(a - 3)^{2}(a + 3)} $．
(2) $ \frac{a - 2}{b(a - 2)^{2}(a + 2)},\frac{a + 2}{b(a - 2)^{2}(a + 2)} $．
(3) $ \frac{y^{2}}{xy(x - y)^{2}},\frac{x^{2}}{xy(x - y)^{2}} $．
(4) $ \frac{5(2x - 3)^{2}}{(2 - 3x)(2 + 3x)(2x - 3)^{2}},\frac{2m(2 - 3x)(2 + 3x)}{(2 - 3x)(2 + 3x)(2x - 3)^{2}} $．
(5) $ \frac{x(x + y)^{2}}{(x - y)(x + y)^{2}},\frac{x(x - y)}{(x - y)(x + y)^{2}} $，
$ \frac{-2(x + y)}{(x - y)(x + y)^{2}} $．
(6) $ \frac{-2(a - 3)(a + 3)}{3(a - 3)^{2}(a + 3)},\frac{3(a + 3)}{3(a - 3)^{2}(a + 3)} $，
$ \frac{2(a - 3)}{3(a - 3)^{2}(a + 3)} $．
12. 已知$\frac{3x - 2}{x^{2}-1}=\frac{A}{x - 1}+\frac{B}{x + 1}$，求$A,B$的值.
答案
12. 解：$ \because \frac{A}{x - 1}+\frac{B}{x + 1}=\frac{A(x + 1)+B(x - 1)}{(x + 1)(x - 1)}=\frac{Ax + A + Bx - B}{(x + 1)(x - 1)}=\frac{(A + B)x + A - B}{x^{2} - 1} $．
$ \because \frac{3x - 2}{x^{2} - 1}=\frac{A}{x - 1}+\frac{B}{x + 1} $，
$ \therefore \frac{3x - 2}{x^{2} - 1}=\frac{(A + B)x + A - B}{x^{2} - 1} $．
$ \therefore \begin{cases}A + B = 3,\\A - B = -2,\end{cases} $ 解得 $ \begin{cases}A = \frac{1}{2},\\B = \frac{5}{2}.\end{cases} $
$ \because \frac{3x - 2}{x^{2} - 1}=\frac{A}{x - 1}+\frac{B}{x + 1} $，
$ \therefore \frac{3x - 2}{x^{2} - 1}=\frac{(A + B)x + A - B}{x^{2} - 1} $．
$ \therefore \begin{cases}A + B = 3,\\A - B = -2,\end{cases} $ 解得 $ \begin{cases}A = \frac{1}{2},\\B = \frac{5}{2}.\end{cases} $
