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6. 已知$x^{2}-x - 1 = 0$，计算$(\dfrac{2}{x + 1}-\dfrac{1}{x})÷\dfrac{x^{2}-x}{x^{2}+2x + 1}$的值是()
A.$1$
B.$-1$
C.$2$
D.$-2$
A.$1$
B.$-1$
C.$2$
D.$-2$
答案
A
解析
先计算括号内：$\dfrac{2}{x + 1} - \dfrac{1}{x} = \dfrac{2x - (x + 1)}{x(x + 1)} = \dfrac{x - 1}{x(x + 1)}$；再将除法转化为乘法：$\dfrac{x - 1}{x(x + 1)} ÷ \dfrac{x^2 - x}{x^2 + 2x + 1} = \dfrac{x - 1}{x(x + 1)} × \dfrac{(x + 1)^2}{x(x - 1)}$；约分后得$\dfrac{x + 1}{x^2}$。由$x^2 - x - 1 = 0$得$x^2 = x + 1$，代入得$\dfrac{x + 1}{x + 1} = 1$。
7. 已知$2a^{2}-7 = 2a$，则代数式$(a-\dfrac{2a - 1}{a})÷\dfrac{a - 1}{a^{2}}$的值为.
答案
$\dfrac{7}{2}$
解析
先化简代数式：
括号内通分：$a - \dfrac{2a - 1}{a} = \dfrac{a^2 - (2a - 1)}{a} = \dfrac{a^2 - 2a + 1}{a} = \dfrac{(a - 1)^2}{a}$；
除法变乘法：$\dfrac{(a - 1)^2}{a} ÷ \dfrac{a - 1}{a^2} = \dfrac{(a - 1)^2}{a} · \dfrac{a^2}{a - 1} = a(a - 1) = a^2 - a$；
由已知$2a^2 - 7 = 2a$，整理得$2a^2 - 2a = 7$，两边同除以2得$a^2 - a = \dfrac{7}{2}$。
括号内通分：$a - \dfrac{2a - 1}{a} = \dfrac{a^2 - (2a - 1)}{a} = \dfrac{a^2 - 2a + 1}{a} = \dfrac{(a - 1)^2}{a}$；
除法变乘法：$\dfrac{(a - 1)^2}{a} ÷ \dfrac{a - 1}{a^2} = \dfrac{(a - 1)^2}{a} · \dfrac{a^2}{a - 1} = a(a - 1) = a^2 - a$；
由已知$2a^2 - 7 = 2a$，整理得$2a^2 - 2a = 7$，两边同除以2得$a^2 - a = \dfrac{7}{2}$。
8. (2025·杭州期中)对于正数$x$，规定$f(x)=\dfrac{x - 1}{x}$，则$f(\dfrac{2026}{2025})+··· +f(\dfrac{6}{5})+f(\dfrac{4}{3})+f(2)+f(4)+f(6)+··· +f(2026)$的值为.
答案
$\frac{2025}{2}$
解析
由题意，$f(x)=\frac{x-1}{x}=1-\frac{1}{x}$。
观察第一部分：$f(\frac{2026}{2025})+···+f(\frac{6}{5})+f(\frac{4}{3})+f(2)$，自变量为$\frac{2k}{2k-1}(k=1,2,···,1013)$，即$\frac{2}{1},\frac{4}{3},···,\frac{2026}{2025}$，共1013项；
第二部分：$f(4)+f(6)+···+f(2026)$，自变量为$2m(m=2,3,···,1013)$，即$4,6,···,2026$，共1012项。
对$m=2,3,···,1013$，有$f(\frac{2m}{2m-1})+f(2m)=\frac{\frac{2m}{2m-1}-1}{\frac{2m}{2m-1}}+\frac{2m-1}{2m}=\frac{1}{2m}+\frac{2m-1}{2m}=1$，共1012对，和为$1012×1=1012$。
第一部分剩余项为$f(2)=\frac{2-1}{2}=\frac{1}{2}$。
总和为$1012+\frac{1}{2}=\frac{2025}{2}$。
观察第一部分：$f(\frac{2026}{2025})+···+f(\frac{6}{5})+f(\frac{4}{3})+f(2)$，自变量为$\frac{2k}{2k-1}(k=1,2,···,1013)$，即$\frac{2}{1},\frac{4}{3},···,\frac{2026}{2025}$，共1013项；
第二部分：$f(4)+f(6)+···+f(2026)$，自变量为$2m(m=2,3,···,1013)$，即$4,6,···,2026$，共1012项。
对$m=2,3,···,1013$，有$f(\frac{2m}{2m-1})+f(2m)=\frac{\frac{2m}{2m-1}-1}{\frac{2m}{2m-1}}+\frac{2m-1}{2m}=\frac{1}{2m}+\frac{2m-1}{2m}=1$，共1012对，和为$1012×1=1012$。
第一部分剩余项为$f(2)=\frac{2-1}{2}=\frac{1}{2}$。
总和为$1012+\frac{1}{2}=\frac{2025}{2}$。
9. 计算：
(1)$(\dfrac{y}{-3x})^{3}·\dfrac{x}{y^{2}}÷(-\dfrac{y}{x})^{4}$；
(2)$(\dfrac{8}{a + 3}+a - 3)÷\dfrac{a^{2}+2a + 1}{a + 3}$
(1)$(\dfrac{y}{-3x})^{3}·\dfrac{x}{y^{2}}÷(-\dfrac{y}{x})^{4}$；
(2)$(\dfrac{8}{a + 3}+a - 3)÷\dfrac{a^{2}+2a + 1}{a + 3}$
答案
(1)
$\begin{aligned}(\dfrac{y}{-3x})^{3}·\dfrac{x}{y^{2}}÷(-\dfrac{y}{x})^{4} =\dfrac{y^{3}}{-27x^{3}}·\dfrac{x}{y^{2}}÷\dfrac{y^{4}}{x^{4}} \\= \dfrac{y^{3}}{-27x^{3}}·\dfrac{x}{y^{2}}·\dfrac{x^{4}}{y^{4}} \\= -\dfrac{x^{2}}{27y^{3}}\end{aligned}$
(2)
$\begin{aligned}(\dfrac{8}{a + 3}+a - 3)÷\dfrac{a^{2}+2a + 1}{a + 3} = \dfrac{8 + (a - 3)(a + 3)}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{8 + a^{2}-9}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{a^{2}-1}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{(a + 1)(a - 1)}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{a - 1}{a + 1}\end{aligned}$
$\begin{aligned}(\dfrac{y}{-3x})^{3}·\dfrac{x}{y^{2}}÷(-\dfrac{y}{x})^{4} =\dfrac{y^{3}}{-27x^{3}}·\dfrac{x}{y^{2}}÷\dfrac{y^{4}}{x^{4}} \\= \dfrac{y^{3}}{-27x^{3}}·\dfrac{x}{y^{2}}·\dfrac{x^{4}}{y^{4}} \\= -\dfrac{x^{2}}{27y^{3}}\end{aligned}$
(2)
$\begin{aligned}(\dfrac{8}{a + 3}+a - 3)÷\dfrac{a^{2}+2a + 1}{a + 3} = \dfrac{8 + (a - 3)(a + 3)}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{8 + a^{2}-9}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{a^{2}-1}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{(a + 1)(a - 1)}{a + 3}·\dfrac{a + 3}{(a + 1)^{2}} \\= \dfrac{a - 1}{a + 1}\end{aligned}$
10. 先化简，再求值：
(1)$(1-\dfrac{1}{a + 2})÷\dfrac{a^{2}-1}{a + 2}$，其中$a = 3$；
(2)$(\dfrac{a}{a - b}-\dfrac{a^{2}}{a^{2}-2ab + b^{2}})÷(\dfrac{a}{a + b}-\dfrac{a^{2}}{a^{2}-b^{2}})+1$，其中$a=\dfrac{2}{3}$，$b = - 3$.
(1)$(1-\dfrac{1}{a + 2})÷\dfrac{a^{2}-1}{a + 2}$，其中$a = 3$；
(2)$(\dfrac{a}{a - b}-\dfrac{a^{2}}{a^{2}-2ab + b^{2}})÷(\dfrac{a}{a + b}-\dfrac{a^{2}}{a^{2}-b^{2}})+1$，其中$a=\dfrac{2}{3}$，$b = - 3$.
答案
(1) 原式$=(1-\dfrac{1}{a + 2})÷\dfrac{a^{2}-1}{a + 2}$
$=\dfrac{a + 2 - 1}{a + 2}×\dfrac{a + 2}{(a - 1)(a + 1)}$
$=\dfrac{a + 1}{a + 2}×\dfrac{a + 2}{(a - 1)(a + 1)}$
$=\dfrac{1}{a - 1}$
当$a = 3$时，原式$=\dfrac{1}{3 - 1}=\dfrac{1}{2}$
(2) 原式$=(\dfrac{a}{a - b}-\dfrac{a^{2}}{(a - b)^{2}})÷(\dfrac{a}{a + b}-\dfrac{a^{2}}{(a + b)(a - b)}) + 1$
$=\dfrac{a(a - b)-a^{2}}{(a - b)^{2}}÷\dfrac{a(a - b)-a^{2}}{(a + b)(a - b)} + 1$
$=\dfrac{-ab}{(a - b)^{2}}÷\dfrac{-ab}{(a + b)(a - b)} + 1$
$=\dfrac{-ab}{(a - b)^{2}}×\dfrac{(a + b)(a - b)}{-ab} + 1$
$=\dfrac{a + b}{a - b} + 1$
$=\dfrac{a + b + a - b}{a - b}$
$=\dfrac{2a}{a - b}$
当$a=\dfrac{2}{3}$，$b = - 3$时，原式$=\dfrac{2×\dfrac{2}{3}}{\dfrac{2}{3}-(-3)}=\dfrac{\dfrac{4}{3}}{\dfrac{11}{3}}=\dfrac{4}{11}$
$=\dfrac{a + 2 - 1}{a + 2}×\dfrac{a + 2}{(a - 1)(a + 1)}$
$=\dfrac{a + 1}{a + 2}×\dfrac{a + 2}{(a - 1)(a + 1)}$
$=\dfrac{1}{a - 1}$
当$a = 3$时，原式$=\dfrac{1}{3 - 1}=\dfrac{1}{2}$
(2) 原式$=(\dfrac{a}{a - b}-\dfrac{a^{2}}{(a - b)^{2}})÷(\dfrac{a}{a + b}-\dfrac{a^{2}}{(a + b)(a - b)}) + 1$
$=\dfrac{a(a - b)-a^{2}}{(a - b)^{2}}÷\dfrac{a(a - b)-a^{2}}{(a + b)(a - b)} + 1$
$=\dfrac{-ab}{(a - b)^{2}}÷\dfrac{-ab}{(a + b)(a - b)} + 1$
$=\dfrac{-ab}{(a - b)^{2}}×\dfrac{(a + b)(a - b)}{-ab} + 1$
$=\dfrac{a + b}{a - b} + 1$
$=\dfrac{a + b + a - b}{a - b}$
$=\dfrac{2a}{a - b}$
当$a=\dfrac{2}{3}$，$b = - 3$时，原式$=\dfrac{2×\dfrac{2}{3}}{\dfrac{2}{3}-(-3)}=\dfrac{\dfrac{4}{3}}{\dfrac{11}{3}}=\dfrac{4}{11}$
11. (2025·遂宁)先化简，再求值：$(a + 1+\dfrac{1}{a - 1})÷\dfrac{a^{3}-2a^{2}}{a^{2}-4a + 4}$，其中$a$满足$a^{2}-4 = 0$.
答案
$\dfrac{4}{3}$
解析
1. 化简原式：
$\begin{aligned}&(a + 1+\dfrac{1}{a - 1})÷\dfrac{a^{3}-2a^{2}}{a^{2}-4a + 4}\\=&(\dfrac{(a + 1)(a - 1)}{a - 1}+\dfrac{1}{a - 1})÷\dfrac{a^{2}(a - 2)}{(a - 2)^2}\\=&\dfrac{a^2 - 1 + 1}{a - 1}×\dfrac{(a - 2)^2}{a^2(a - 2)}\\=&\dfrac{a^2}{a - 1}×\dfrac{a - 2}{a^2}\\=&\dfrac{a - 2}{a - 1}\end{aligned}$
2. 由$a^2 - 4 = 0$得$a = ±2$，又因分母不为0，$a ≠ 1,0,2$，故$a = -2$。
3. 代入$a = -2$：$\dfrac{-2 - 2}{-2 - 1}=\dfrac{-4}{-3}=\dfrac{4}{3}$
$\begin{aligned}&(a + 1+\dfrac{1}{a - 1})÷\dfrac{a^{3}-2a^{2}}{a^{2}-4a + 4}\\=&(\dfrac{(a + 1)(a - 1)}{a - 1}+\dfrac{1}{a - 1})÷\dfrac{a^{2}(a - 2)}{(a - 2)^2}\\=&\dfrac{a^2 - 1 + 1}{a - 1}×\dfrac{(a - 2)^2}{a^2(a - 2)}\\=&\dfrac{a^2}{a - 1}×\dfrac{a - 2}{a^2}\\=&\dfrac{a - 2}{a - 1}\end{aligned}$
2. 由$a^2 - 4 = 0$得$a = ±2$，又因分母不为0，$a ≠ 1,0,2$，故$a = -2$。
3. 代入$a = -2$：$\dfrac{-2 - 2}{-2 - 1}=\dfrac{-4}{-3}=\dfrac{4}{3}$
