试题:
计算
(1)
2
x+1
-
1
x-1

(2)0.25×(-2)-2÷(16)-1-(π-3)0
(3)
a-1
a+2
a2-4
a2-2a+1
÷
1
a2-1

答案:

我来补答
(1)
2
x+1
-
1
x-1
=
2(x-1)
(x+1)(x-1)
-
x+1
(x+1)(x-1)
=
x-3
(x+1)(x-1)

(2)0.25×(-2)-2÷(16)-1-(π-3)0=
1
4
×
1
4
÷
1
16
-1=
1
16
×16-1=0;
(3)
a-1
a+2
a2-4
a2-2a+1
÷
1
a2-1
=
a-1
a+2
(a+2)(a-2)
(a-1)2
×(a+1)(a-1)=(a-2)(a+1)=a2-a-2.
 
 
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