《2026年练就优等生课后提分攻略八年级数学全一册人教版》

01 (1) 若$\frac{1}{m}+\frac{1}{n}=2$，则分式$\frac{5m+5n-2mn}{-m-n}$的值为-4；
(2) 已知$a+b+c=0$，$abc\neq0$，求$\frac{1}{a^{2}+b^{2}-c^{2}}+\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}$的值.

答案： 01 解：
(1)-4
(2)
∵$a + b + c = 0$,
∴$a + b = -c$,$b + c = -a$,$a + c = -b$.
∴原式$ = \frac{1}{a^{2} + b^{2} - (a + b)^{2}} + \frac{1}{b^{2} + c^{2} - (b + c)^{2}} +$
$\frac{1}{c^{2} + a^{2} - (c + a)^{2}} = - \frac{a + b + c}{2abc} = 0$.

02 (1) 已知$a+\frac{1}{a}=4$，求下列各式的值.
①$a^{2}+\frac{1}{a^{2}}$；②$\frac{a^{2}}{a^{4}+a^{2}+1}$.
(2) 已知$\frac{ab}{a+b}=\frac{1}{3}$，$\frac{bc}{b+c}=\frac{1}{4}$，$\frac{ac}{a+c}=\frac{1}{5}$，求$\frac{abc}{ab+bc+ca}$的值.

答案： 02 解：
(1)①
∵$a + \frac{1}{a} = 4$,
∴$(a + \frac{1}{a})^{2} = 4^{2}$，
即$a^{2} + 2 + \frac{1}{a^{2}} = 16$.
∴$a^{2} + \frac{1}{a^{2}} = 14$.
②
∵$\frac{a^{4} + a^{2} + 1}{a^{2}} = \frac{a^{4}}{a^{2}} + \frac{a^{2}}{a^{2}} + \frac{1}{a^{2}} = a^{2} + 1 + \frac{1}{a^{2}} = 14 + 1 = 15$，
∴$\frac{a^{2}}{a^{4} + a^{2} + 1} = \frac{1}{15}$.
(2)
∵$\frac{ab}{a + b} = \frac{1}{3}$,
∴$\frac{a + b}{ab} = \frac{1}{b} + \frac{1}{a} = 3$.
同理可得$\frac{1}{b} + \frac{1}{c} = 4$,$\frac{1}{a} + \frac{1}{c} = 5$，
∴$\frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{c} = 3 + 4 + 5 = 12$.
∴$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 6$.
原式$ =\frac{1}{ \frac{1}{c} + \frac{1}{a} + \frac{1}{b}}= \frac{1}{6}$.

03 (1) 已知$3x=2y=5z\neq0$，求$\frac{x+2y+3z}{x-y+z}$的值；
(2) 已知$\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}$，求$\frac{abc}{(a+b)(b+c)(c+a)}$的值.

答案： 03 解：
(1)设$3x = 2y = 5z = 30k(k \neq 0)$，则$x = 10k$,$y = 15k$,$z = 6k$.
原式$ = \frac{10k + 30k + 18k}{10k - 15k + 6k} = \frac{58k}{k} = 58$.
(2)设$\frac{b + c}{a} = \frac{c + a}{b} = \frac{a + b}{c} = k$，
∴$b + c = ak$,$c + a = bk$,$a + b = ck$.
∴$b + c + c + a + a + b = ak + bk + ck$,$2(a + b + c) = k(a + b + c)$,即$(a + b + c)(2 - k) = 0$，
∴$a + b + c = 0$或$k = 2$.
由$a + b + c = 0$时，$\frac{-a}{a} = \frac{-b}{b} = \frac{-c}{c} = k$，得$k = -1$.
∴$k = 2$或$k = -1$.
∴原式$ = \frac{abc}{abck^{3}} = \frac{1}{k^{3}}$.
∴$\frac{abc}{(a + b)(b + c)(c + a)}$的值为$\frac{1}{8}$或$-1$.