先化简再求值:(1)2xx-3+x3-x,其中x=-1;(2)(3aa-1-aa+1)•a2-1a,其中a=2.(3)(1+1a2-1)÷aa-1,其中a=-3.(4)(1a-2+1a+2)÷2aa2

题目简介

先化简再求值:(1)2xx-3+x3-x,其中x=-1;(2)(3aa-1-aa+1)•a2-1a,其中a=2.(3)(1+1a2-1)÷aa-1,其中a=-3.(4)(1a-2+1a+2)÷2aa2

题目详情

先化简再求值:
(1)
2x
x-3
+
x
3-x
,其中x=-1;        
(2)(
3a
a-1
-
a
a+1
)•
a2-1
a
,其中a=2.
(3)(1+
1
a2-1
a
a-1
,其中a=-3.   
(4)(
1
a-2
+
1
a+2
2a
a2-4a+4
,其中a=-4.
题型:解答题难度:中档来源:不详

答案

(1)原式=class="stub"2x-x
x-3
=class="stub"x
x-3
,当x=-1时,原式=class="stub"-1
-1-3
=class="stub"1
4

(2)原式=(class="stub"3a
a-1
-
class="stub"a
a+1
)•
(a+1)(a-1)
a
=class="stub"3a
a-1
(a+1)(a-1)
a
-
class="stub"a
a+1
(a+1)(a-1)
a
=3(a+1)-(a-1)=2a+4,当a=2时,原式=4+4=8;
(3)原式=[1+class="stub"1
(a+1)(a-1)
]•class="stub"a-1
a
=class="stub"a-1
a
+class="stub"1
a(a+1)
=
(a-1)(a+1)+1
a(a+1)
=
a2
a(a+1)
=class="stub"a
a+1
,当a=-3时,原式=class="stub"-3
-2
=class="stub"3
2

(4)原式=(class="stub"1
a-2
+class="stub"1
a+2
)•
(a-2)2
2a
=class="stub"1
a-2
(a-2)2
2a
+class="stub"1
a+2
(a-2)2
2a
=class="stub"a-2
2a
+
(a-2)2
2a(a+2)
=
(a-2)(a+2)+(a-2)2
2a(a+2)
=
2a2-4a
2a(a+2)
=class="stub"a-2
a+2
,当a=-4时,原式=class="stub"-6
-2
=3.

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