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1[2026山东淄博期末]一个水池有甲、乙两个进水管，单独开甲、乙管各需要$x$小时、$y$小时可注满水池，现两管同时打开，则注满水池的时间为（
A.$\frac{x+y}{xy}$小时
B.$\frac{xy}{x+y}$小时
C.$\frac{1}{x+y}$小时
D.$\frac{1}{xy}$小时
B
）A.$\frac{x+y}{xy}$小时
B.$\frac{xy}{x+y}$小时
C.$\frac{1}{x+y}$小时
D.$\frac{1}{xy}$小时
答案
1.B 【解析】设水池注满后总水量为 1,则甲、乙管的工作效率分别为$\frac{1}{x},\frac{1}{y}$,$\therefore$ 两管同时打开,注满水池的时间为 $1 ÷ (\frac{1}{x}+\frac{1}{y}) =\frac{xy}{x+y}$(时).故选 B.
2[2026甘肃平凉质检]老师在黑板上给出了一道分式计算题：$\frac{x}{x^2 -1} ÷ ( \frac{1}{x-1} + \frac{1}{x+1} )$。沙沙解答过程：$\frac{x}{x^2 -1} ÷ ( \frac{1}{x-1} + \frac{1}{x+1} ) = \frac{x}{(x+1)(x-1)} × (x-1) + \frac{x}{(x+1)(x-1)} × (x+1) ··· ①$
$= \frac{x}{x+1} + \frac{x}{x-1} ··· ②$
$= \frac{2x^2}{(x+1)(x-1)} ··· ③$
沙沙的解答过程是从
A.①，$\frac{1}{2}$
B.②，$\frac{1}{2}$
C.②，$-\frac{1}{2}$
D.①，$-\frac{1}{2}$
$= \frac{x}{x+1} + \frac{x}{x-1} ··· ②$
$= \frac{2x^2}{(x+1)(x-1)} ··· ③$
沙沙的解答过程是从
①
开始出现错误的，正确的结果是$\frac{1}{2}$
，下列填在横线上的答案正确的是（A
）A.①，$\frac{1}{2}$
B.②，$\frac{1}{2}$
C.②，$-\frac{1}{2}$
D.①，$-\frac{1}{2}$
答案
2.A 【解析】沙沙的解答过程是从①开始出现错误的. 正确的解答过程如下: 原式 = $\frac{x}{(x+1)(x-1)} ÷ \frac{x+1+x-1}{(x+1)(x-1)} = \frac{x}{(x+1)(x-1)} × \frac{(x+1)(x-1)}{2x} = \frac{1}{2}$, 则正确的结果是 $\frac{1}{2}$, 故选 A.
3[2026河北秦皇岛期末]若$(□ -1)× \dfrac{1}{5-x}=\dfrac{1}{x-4}$，则“□”表示的最简分式为
$\frac{1}{x-4}$
。答案
3.$\frac{1}{x-4}$ 【解析】根据等式的性质可得$□=\frac{1}{x-4}÷\frac{1}{5-x}+1 = \frac{5-x}{x-4}+1 = \frac{5-x+x-4}{x-4} = \frac{1}{x-4}$. 故答案为$\frac{1}{x-4}$.
4 已知$\dfrac{3-x}{x^2-2x+1}÷ P=1+\dfrac{x^2-2x-1}{1-x}.$
(1)求$P$(化成最简形式)；
(2)当$x=n$时,记$P$的值为$P(n)$.如:当$x=2$时,$P$的值为$P(2)$；当$x=3$时,$P$的值为$P(3)$；…请写出关于$t$的不等式$\dfrac{t-2}{4}-\dfrac{3-t}{2}≥ P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)$的解集及最小整数解.
(1)求$P$(化成最简形式)；
(2)当$x=n$时,记$P$的值为$P(n)$.如:当$x=2$时,$P$的值为$P(2)$；当$x=3$时,$P$的值为$P(3)$；…请写出关于$t$的不等式$\dfrac{t-2}{4}-\dfrac{3-t}{2}≥ P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)$的解集及最小整数解.
答案
4.【解】(1) $P = \frac{3-x}{x^2-2x+1} ÷ ( 1 + \frac{x^2-2x-1}{1-x} ) = \frac{3-x}{(x-1)^2} ÷ \frac{1-x+x^2-2x-1}{1-x} = \frac{3-x}{(x-1)^2} · \frac{1-x}{x(x-3)} = \frac{1}{x(x-1)} = \frac{1}{x^2-x}$.
(2)$P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8) = \frac{1}{2×(2-1)} + \frac{1}{3×(3-1)} + \frac{1}{4×(4-1)} + \frac{1}{5×(5-1)} + \frac{1}{6×(6-1)} + \frac{1}{7×(7-1)} + \frac{1}{8×(8-1)} = 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8} = 1-\frac{1}{8} = \frac{7}{8}$,$\therefore \frac{t-2}{4}-\frac{3-t}{2} ≥ \frac{7}{8}$,
$\therefore 2(t-2)-4(3-t)≥7$,解得 $t≥3\frac{5}{6}$,$\therefore$ 解集为$t≥3\frac{5}{6}$,最小整数解为4.
思路分析
(2)先求出$P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)$的值,再解不等式即可.
(2)$P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8) = \frac{1}{2×(2-1)} + \frac{1}{3×(3-1)} + \frac{1}{4×(4-1)} + \frac{1}{5×(5-1)} + \frac{1}{6×(6-1)} + \frac{1}{7×(7-1)} + \frac{1}{8×(8-1)} = 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8} = 1-\frac{1}{8} = \frac{7}{8}$,$\therefore \frac{t-2}{4}-\frac{3-t}{2} ≥ \frac{7}{8}$,
$\therefore 2(t-2)-4(3-t)≥7$,解得 $t≥3\frac{5}{6}$,$\therefore$ 解集为$t≥3\frac{5}{6}$,最小整数解为4.
思路分析
(2)先求出$P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)$的值,再解不等式即可.
5 已知 $ a - 2b = 0 $ 且 $ b ≠ 0 $，则 $ ( \frac{b}{a - b} + 1 ) · \frac{a^2 - b^2}{a^2} $ 的值为（
A.$ \frac{3}{2} $
B.$ \frac{1}{2} $
C.3
D.-1
A
）A.$ \frac{3}{2} $
B.$ \frac{1}{2} $
C.3
D.-1
答案
5.A 【解析】原式$=(\frac{b}{a-b}+\frac{a-b}{a-b}) · \frac{(a+b)(a-b)}{a^2} = \frac{a}{a-b} · \frac{(a+b)(a-b)}{a^2} = \frac{a+b}{a}$. $\because a-2b=0,b≠0$,
$\therefore a=2b$,则原式$=\frac{2b+b}{2b}=\frac{3}{2}$,故选 A.
$\therefore a=2b$,则原式$=\frac{2b+b}{2b}=\frac{3}{2}$,故选 A.
6 有这样一道题:计算$\frac{x^2 - 2x + 1}{x^2 - 1} ÷ \frac{x - 1}{x^2 + x} - x$的值,其中$x=2004$. 甲同学把“$x=2004$”错抄成“$x=2040$”,但他的计算结果是正确的,请说明原因.
答案
6.【解】$\because \frac{x^2-2x+1}{x^2-1} ÷ \frac{x-1}{x^2+x} -x = \frac{(x-1)^2}{(x-1)(x+1)} · \frac{x(x+1)}{x-1} -x = x-x=0$,$\therefore$ 无论$x$为何值,原式的值都为0,$\therefore$ 甲同学把“$x=2 004$”错抄成“$x=2 040$”,其结果也是0,$\therefore$ 他的计算结果是正确的.
方法归纳
分式的化简求值一般先把分式化简,再根据已知条件代入求值,当需要自主选择字母的取值时,要注意所选取的字母的值要使分式有意义.
方法归纳
分式的化简求值一般先把分式化简,再根据已知条件代入求值,当需要自主选择字母的取值时,要注意所选取的字母的值要使分式有意义.
7[2025安徽池州期末]先化简：$\frac{x^2 -4x +4}{x+1} ÷ ( \frac{3}{x+1} - x+1 )$，再从不等式组$\begin{cases}5-2x≥1, \\x+1＞0 \end{cases}$的整数解中选择一个合适的值代入求值.
刷素养 走向重高
刷素养 走向重高
答案
7.【解】$\frac{x^2-4x+4}{x+1} ÷ ( \frac{3}{x+1}-x+1 ) = \frac{(x-2)^2}{x+1} ÷ [ \frac{3}{x+1}-\frac{(x-1)(x+1)}{x+1} ] = \frac{(x-2)^2}{x+1} ÷ \frac{4-x^2}{x+1} = \frac{(2-x)^2}{x+1} · \frac{x+1}{(2+x)(2-x)} = \frac{2-x}{2+x}$. 解不等式组$\begin{cases}5-2x≥1, \\x+1＞0,\end{cases}$
得$-1＜x≤2$,$\therefore$ 不等式组的整数解为0,1,2.
$\because x+1≠0,(2+x)(2-x)≠0$,$\therefore x≠-1$,且$x≠±2$,$\therefore x=0$或1. 当$x=1$时,原式$=\frac{2-1}{2+1}=\frac{1}{3}$.
(或当$x=0$时,原式$=\frac{2-0}{2+0}=1$)
得$-1＜x≤2$,$\therefore$ 不等式组的整数解为0,1,2.
$\because x+1≠0,(2+x)(2-x)≠0$,$\therefore x≠-1$,且$x≠±2$,$\therefore x=0$或1. 当$x=1$时,原式$=\frac{2-1}{2+1}=\frac{1}{3}$.
(或当$x=0$时,原式$=\frac{2-0}{2+0}=1$)
8 核心素养运算能力 [中]已知$a,b,c$是非零有理数，且满足$ab^2 = \frac{c}{a} - b$，则
$( \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} ) ÷ ( \frac{2}{ab} - \frac{2ab}{c} ) ÷ \frac{101}{c}$
等于
$( \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} ) ÷ ( \frac{2}{ab} - \frac{2ab}{c} ) ÷ \frac{101}{c}$
等于
$-\frac{1}{202}$
。答案
8.$-\frac{1}{202}$ 【解析】$\because ab^2 = \frac{c}{a} - b$,$\therefore a^2b^2 = c - ab$,
$a^2b^2 - c = -ab$,$c - a^2b^2 = ab$,$\therefore \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} = ( \frac{ab}{c} - \frac{1}{ab} )^2 + \frac{2a^2b^2}{abc^2} - \frac{2c}{abc^2} = ( \frac{a^2b^2 - c}{abc} )^2 + \frac{2(a^2b^2 - c)}{abc^2} = ( \frac{-ab}{abc} )^2 + \frac{-2ab}{abc^2} = \frac{1}{c^2} - \frac{2}{c^2} = -\frac{1}{c^2}$,$\frac{2}{ab} - \frac{2ab}{c} = \frac{2c-2a^2b^2}{abc} = \frac{2ab}{abc} = \frac{2}{c}$,
$\therefore ( \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} ) ÷ ( \frac{2}{ab} - \frac{2ab}{c} ) ÷ \frac{101}{c} = -\frac{1}{c^2} ÷ \frac{2}{c} ÷ \frac{101}{c} = -\frac{1}{c^2} · \frac{c}{2} · \frac{c}{101} = -\frac{1}{202}$.
故答案为$-\frac{1}{202}$.
$a^2b^2 - c = -ab$,$c - a^2b^2 = ab$,$\therefore \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} = ( \frac{ab}{c} - \frac{1}{ab} )^2 + \frac{2a^2b^2}{abc^2} - \frac{2c}{abc^2} = ( \frac{a^2b^2 - c}{abc} )^2 + \frac{2(a^2b^2 - c)}{abc^2} = ( \frac{-ab}{abc} )^2 + \frac{-2ab}{abc^2} = \frac{1}{c^2} - \frac{2}{c^2} = -\frac{1}{c^2}$,$\frac{2}{ab} - \frac{2ab}{c} = \frac{2c-2a^2b^2}{abc} = \frac{2ab}{abc} = \frac{2}{c}$,
$\therefore ( \frac{a^2b^2}{c^2} - \frac{2}{c} + \frac{1}{a^2b^2} + \frac{2ab}{c^2} - \frac{2}{abc} ) ÷ ( \frac{2}{ab} - \frac{2ab}{c} ) ÷ \frac{101}{c} = -\frac{1}{c^2} ÷ \frac{2}{c} ÷ \frac{101}{c} = -\frac{1}{c^2} · \frac{c}{2} · \frac{c}{101} = -\frac{1}{202}$.
故答案为$-\frac{1}{202}$.
