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20.(本小题5分)先化简,再求值:$\frac{x^2 -1}{x^2 -x} ÷ (x + \frac{2x +1}{x})$,其中$x=2.$
答案
20. 【解】$\frac{x^2-1}{x^2-x} ÷ (x+\frac{2x+1}{x})=\frac{x^2-1}{x^2-x} ÷ \frac{x^2+2x+1}{x}$
$=\frac{(x+1)(x-1)}{x(x-1)} ÷ \frac{(x+1)^2}{x}$
$=\frac{(x+1)(x-1)}{x(x-1)} · \frac{x}{(x+1)^2}=\frac{1}{x+1}$,
当x=2时,原式=$\frac{1}{2+1}=\frac{1}{3}$.
$=\frac{(x+1)(x-1)}{x(x-1)} ÷ \frac{(x+1)^2}{x}$
$=\frac{(x+1)(x-1)}{x(x-1)} · \frac{x}{(x+1)^2}=\frac{1}{x+1}$,
当x=2时,原式=$\frac{1}{2+1}=\frac{1}{3}$.
