2025年自我提升与评价八年级数学上册人教版第151页答案
1. 计算$\left(x-\frac{1}{x}\right)\cdot \frac{x}{x+1}$的结果是(
D
)
A.$\frac{1}{x+1}$
B.$\frac{1}{x-1}$
C.$x+1$
D.$x-1$

答案

D

解析

$\begin{aligned}\left(x - \frac{1}{x}\right) \cdot \frac{x}{x + 1}&=\left(\frac{x^2}{x} - \frac{1}{x}\right) \cdot \frac{x}{x + 1}\\&=\frac{x^2 - 1}{x} \cdot \frac{x}{x + 1}\\&=\frac{(x + 1)(x - 1)}{x} \cdot \frac{x}{x + 1}\\&=x - 1\end{aligned}$
D
2. 计算$\left(a-\frac{1}{b}\right)÷ \left(\frac{1}{a}-b\right)$的结果是(
A
)
A.$-\frac{a}{b}$
B.$\frac{a}{b}$
C.$-\frac{b}{a}$
D.$\frac{b}{a}$

答案

A

解析

$\begin{aligned}\left(a - \frac{1}{b}\right) ÷ \left(\frac{1}{a} - b\right) &= \left(\frac{ab - 1}{b}\right) ÷ \left(\frac{1 - ab}{a}\right) \\&= \frac{ab - 1}{b} × \frac{a}{1 - ab} \\&= \frac{-(1 - ab)}{b} × \frac{a}{1 - ab} \\&= -\frac{a}{b}\end{aligned}$
A
3. 试卷上一个正确的式子$\left(\frac{1}{a+b}+\frac{1}{a-b}\right)÷★= \frac{2}{a+b}$被小颖同学不小心滴上墨汁.被墨汁遮住部分的代数式为(
A
)
A.$\frac{a}{a-b}$
B.$\frac{a-b}{a}$
C.$\frac{a}{a+b}$
D.$\frac{4a}{a^{2}-b^{2}}$

答案

A

解析

设被墨汁遮住部分的代数式为$x$，则$\left(\frac{1}{a+b}+\frac{1}{a-b}\right)÷ x = \frac{2}{a+b}$。
$\begin{aligned}\frac{1}{a+b}+\frac{1}{a-b}&=\frac{(a - b)+(a + b)}{(a + b)(a - b)}\\&=\frac{2a}{(a + b)(a - b)}\end{aligned}$
由$\frac{2a}{(a + b)(a - b)}÷ x=\frac{2}{a + b}$，得$x=\frac{2a}{(a + b)(a - b)}÷\frac{2}{a + b}=\frac{a}{a - b}$
A
4. 如果$a^{2}+2a-1= 0$,那么代数式$\left(a-\frac{4}{a}\right)\cdot \frac{a^{2}}{a-2}$的值是(
C
)
A.$-3$
B.$-1$
C.$1$
D.$3$

答案

C

解析

$\begin{aligned}\left(a - \frac{4}{a}\right) \cdot \frac{a^2}{a - 2}&=\left(\frac{a^2 - 4}{a}\right) \cdot \frac{a^2}{a - 2}\\&=\frac{(a + 2)(a - 2)}{a} \cdot \frac{a^2}{a - 2}\\&=a(a + 2)\\&=a^2 + 2a\end{aligned}$
因为$a^2 + 2a - 1 = 0$，所以$a^2 + 2a = 1$，原式的值为$1$。
C
5. 计算:$\frac{2a}{a^{2}-4}\cdot \frac{a-2}{a}+\frac{a}{a+2}=$
1
.

答案

$1$

解析

$\frac{2a}{a^{2}-4}\cdot \frac{a-2}{a}+\frac{a}{a+2}$
$=\frac{2a}{(a+2)(a-2)}\cdot \frac{a-2}{a}+\frac{a}{a+2}$
$=\frac{2}{a+2}+\frac{a}{a+2}$
$=\frac{a+2}{a+2}$
$=1$
6. 计算:$\frac{x-y}{x}÷ \left(x+\frac{y^{2}-2xy}{x}\right)= $
$\frac{1}{x-y}$
.

答案

【解析】：原式$=\frac{x-y}{x}÷\left(\frac{x^{2}}{x}+\frac{y^{2}-2xy}{x}\right)=\frac{x-y}{x}÷\frac{x^{2}-2xy+y^{2}}{x}=\frac{x-y}{x}×\frac{x}{(x-y)^{2}}=\frac{1}{x-y}$
【答案】：$\frac{1}{x-y}$
7. 计算:$\frac{2}{a^{2}-1}÷ \left(\frac{1}{a-1}-\frac{1}{a+1}\right)= $
1
.

答案

1

解析

$\begin{aligned}&\frac{2}{a^{2}-1}÷ \left(\frac{1}{a-1}-\frac{1}{a+1}\right)\\=&\frac{2}{(a-1)(a+1)}÷\left[\frac{a+1-(a-1)}{(a-1)(a+1)}\right]\\=&\frac{2}{(a-1)(a+1)}÷\frac{a+1 - a + 1}{(a-1)(a+1)}\\=&\frac{2}{(a-1)(a+1)}÷\frac{2}{(a-1)(a+1)}\\=&\frac{2}{(a-1)(a+1)}×\frac{(a-1)(a+1)}{2}\\=&1\end{aligned}$
1