问题 解答题
(1)解方程组 
x+y=3
2x-y=6

(2)先化简,再求值:(x+2)(x-2)-x(x-1),其中x=-1.
(3)计算:
a
a-1
÷
a2-a
a2-1
-
1
a-1
答案

(1)方程组

x+y=3①
2x-y=6②

①+②得:3x=9,

解得:x=3,

把x=3代入①得:y=0,

故方程组的解为

x=3
y=0

(2)(x+2)(x-2)-x(x-1)

=x2-4-x2+x

=x-4,

当x=-1时,原式=-1-4=-5;

(3)原式=

a
a-1
(a+1)(a-1)
a(a-1)
-
1
a-1

=

a+1
a-1
-
1
a-1

=

a+1-1
a-1

=

a
a-1

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