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17. (本题满分8分)
计算与求值：
(1)计算：$\frac{x^2-x}{x^2+2x+1}÷\left(\frac{2}{x+1}-\frac{1}{x}\right)$；
(2)先化简$\left(\frac{3}{a+2}+a-2\right)÷\frac{a^2-2a+1}{a+2}$，再从$-2$，$1$，$3$三个数中选取一个合适的数值作为$a$的值代入求值。
计算与求值：
(1)计算：$\frac{x^2-x}{x^2+2x+1}÷\left(\frac{2}{x+1}-\frac{1}{x}\right)$；
(2)先化简$\left(\frac{3}{a+2}+a-2\right)÷\frac{a^2-2a+1}{a+2}$，再从$-2$，$1$，$3$三个数中选取一个合适的数值作为$a$的值代入求值。
答案
(1) $\frac{x^2}{x + 1}$；(2) $\frac{a + 1}{a - 1}$，值为$2$
解析
(1) $\frac{x^2 - x}{x^2 + 2x + 1}÷\left(\frac{2}{x + 1} - \frac{1}{x}\right)$
$\begin{aligned}&=\frac{x(x - 1)}{(x + 1)^2}÷\left(\frac{2x - (x + 1)}{x(x + 1)}\right)\\&=\frac{x(x - 1)}{(x + 1)^2}÷\frac{x - 1}{x(x + 1)}\\&=\frac{x(x - 1)}{(x + 1)^2}×\frac{x(x + 1)}{x - 1}\\&=\frac{x^2}{x + 1}\end{aligned}$
(2) $\left(\frac{3}{a + 2} + a - 2\right)÷\frac{a^2 - 2a + 1}{a + 2}$
$\begin{aligned}&=\left(\frac{3 + (a - 2)(a + 2)}{a + 2}\right)÷\frac{(a - 1)^2}{a + 2}\\&=\frac{a^2 - 1}{a + 2}×\frac{a + 2}{(a - 1)^2}\\&=\frac{(a - 1)(a + 1)}{(a - 1)^2}\\&=\frac{a + 1}{a - 1}\end{aligned}$
$a$不能取$-2$，$1$，取$a = 3$，原式$=\frac{3 + 1}{3 - 1}=2$
$\begin{aligned}&=\frac{x(x - 1)}{(x + 1)^2}÷\left(\frac{2x - (x + 1)}{x(x + 1)}\right)\\&=\frac{x(x - 1)}{(x + 1)^2}÷\frac{x - 1}{x(x + 1)}\\&=\frac{x(x - 1)}{(x + 1)^2}×\frac{x(x + 1)}{x - 1}\\&=\frac{x^2}{x + 1}\end{aligned}$
(2) $\left(\frac{3}{a + 2} + a - 2\right)÷\frac{a^2 - 2a + 1}{a + 2}$
$\begin{aligned}&=\left(\frac{3 + (a - 2)(a + 2)}{a + 2}\right)÷\frac{(a - 1)^2}{a + 2}\\&=\frac{a^2 - 1}{a + 2}×\frac{a + 2}{(a - 1)^2}\\&=\frac{(a - 1)(a + 1)}{(a - 1)^2}\\&=\frac{a + 1}{a - 1}\end{aligned}$
$a$不能取$-2$，$1$，取$a = 3$，原式$=\frac{3 + 1}{3 - 1}=2$
