问题 解答题
计算:
(1)
4
a-2
+
3
2-a
+
1
a2-a-2

(2)
a-4
a2-9
(1+
10a-19
a2-8a+16
)
÷
1
a-3

(3)已知:x2+x-1=0,求x(1-
2
1-x
)
÷(x+1)-
x(x2-1)
x2-2x+1
的值.
答案

(1)原式=

4
a-2
-
3
a-2
+
1
(a-2)(a+1)

=

1
a-2
+
1
(a-2)(a+1)

=

a+1
(a-2)(a+1)
+
1
(a-2)(a+1)

=

a+2
(a-2)(a+1)

(2)原式=

a-4
(a+3)(a-3)
•(
a2-8a+16
(a-4)2
+
10a-19
(a-4)2
)•(a-3)

=

a-4
(a+3)(a-3)
a2+2a-3
(a-4)2
•(a-3)

=

a-1
a-4

(3)原式=

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

=

-x2
x-1

∵x2+x-1=0,

∴x-1=-x2,原式=

-x2
-x2
=1

单项选择题 A1/A2型题
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