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1.(期末·河西区)若$a=1$，则$\frac{a^2}{a+3}-\frac{9}{a+3}$的值为(
A.2
B.$-2$
C.$\frac{1}{2}$
D.$-\frac{1}{2}$
B
)A.2
B.$-2$
C.$\frac{1}{2}$
D.$-\frac{1}{2}$
答案
1. B 【解析】原式=$\frac{a^2-9}{a+3}=\frac{(a+3)(a-3)}{a+3}=a-3$.
当a=1时,原式=1-3=-2.
故选B.
当a=1时,原式=1-3=-2.
故选B.
2.（期末·河北区）先化简,再求值: $(\dfrac{x}{x-1}-\dfrac{x}{x^2-1})÷\dfrac{x^2-x}{x^2-2x+1}$,其中$x=2$.
答案
2.【解】原式=$[\frac{x(x+1)}{(x+1)(x-1)}-\frac{x}{(x+1)(x-1)}]·\frac{x^2-2x+1}{x^2-x}$
$=\frac{x^2+x-x}{(x+1)(x-1)}·\frac{x^2-2x+1}{x^2-x}$
$=\frac{x^2}{(x+1)(x-1)}·\frac{(x-1)^2}{x(x-1)}$
$=\frac{x}{x+1}$.
当x=2时,原式=$\frac{x}{x+1}=\frac{2}{2+1}=\frac{2}{3}$.
$=\frac{x^2+x-x}{(x+1)(x-1)}·\frac{x^2-2x+1}{x^2-x}$
$=\frac{x^2}{(x+1)(x-1)}·\frac{(x-1)^2}{x(x-1)}$
$=\frac{x}{x+1}$.
当x=2时,原式=$\frac{x}{x+1}=\frac{2}{2+1}=\frac{2}{3}$.
3.先化简,再求值: $(\dfrac{x}{x+y}+\dfrac{2y}{x+y})· \dfrac{xy}{x+2y}÷ (\dfrac{1}{x}+\dfrac{1}{y})$,其中$x=3,y=-1.$
答案
3.【解】原式=$\frac{x+2y}{x+y}·\frac{xy}{x+2y}÷\frac{x+y}{xy}$
$=\frac{xy}{x+y}·\frac{xy}{x+y}$
$=(\frac{xy}{x+y})^2$.
当x=3,y=-1时,原式=$[\frac{3×(-1)}{3+(-1)}]^2=(-\frac{3}{2})^2=\frac{9}{4}$.
$=\frac{xy}{x+y}·\frac{xy}{x+y}$
$=(\frac{xy}{x+y})^2$.
当x=3,y=-1时,原式=$[\frac{3×(-1)}{3+(-1)}]^2=(-\frac{3}{2})^2=\frac{9}{4}$.
