计算:(1)4a-2+32-a+1a2-a-2;(2)a-4a2-9(1+10a-19a2-8a+16)÷1a-3;(3)已知:x2+x-1=0,求x(1-21-x)÷(x+1)-x(x2-1)x2-

题目简介

计算:(1)4a-2+32-a+1a2-a-2;(2)a-4a2-9(1+10a-19a2-8a+16)÷1a-3;(3)已知:x2+x-1=0,求x(1-21-x)÷(x+1)-x(x2-1)x2-

题目详情

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.
题型:解答题难度:中档来源:不详

答案

(1)原式=class="stub"4
a-2
-class="stub"3
a-2
+class="stub"1
(a-2)(a+1)

=class="stub"1
a-2
+class="stub"1
(a-2)(a+1)

=class="stub"a+1
(a-2)(a+1)
+class="stub"1
(a-2)(a+1)

=class="stub"a+2
(a-2)(a+1)


(2)原式=class="stub"a-4
(a+3)(a-3)
•(
a2-8a+16
(a-4)2
+class="stub"10a-19
(a-4)2
)•(a-3)
=class="stub"a-4
(a+3)(a-3)
a2+2a-3
(a-4)2
•(a-3)
=class="stub"a-1
a-4


(3)原式=
x(x+1)
x-1
class="stub"1
x+1
-
x(x+1)
x-1

=
-x2
x-1

∵x2+x-1=0,
∴x-1=-x2,原式=
-x2
-x2
=1

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