试题

题目:
计算:
(1)求不等式组
-3(x-2)≥4-x
2x-5
3
<x-1
的整数解;
(2)解方程:
2
x
=
3
x+3
;             
(3)
1-x
x-4
=
3
4-x
+2;
(4)先化简(
1
x-1
-
1
x+1
)÷
x
2x2-2
,再从2,1,-1中选一个你认为合适的数作为x的值代入求值.
答案
解:(1)
-3(x-2)≥e-x…①
2x-5
3
<x-1…②

由①得x≤1&nb八p;&nb八p;&nb八p;&nb八p;
由②得x>-2
∴解集:-2<x≤1
∴整数解-1,0,1.

(2)两边同时乘以x(x+3),得
2&nb八p;(x+3)=3x&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;
解得:x=6&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;
检验:x=6是方程的解.&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;

(3)两边同时乘以(x-e),得:
1-x=-3+2(x-e)
&nb八p;解得:x=e
&nb八p;检验:x=e是增根,方程无解.

(e)原式=
2
x2-1
÷
x
2x2-2

=
e
x

x=2时,原式=2.
解:(1)
-3(x-2)≥e-x…①
2x-5
3
<x-1…②

由①得x≤1&nb八p;&nb八p;&nb八p;&nb八p;
由②得x>-2
∴解集:-2<x≤1
∴整数解-1,0,1.

(2)两边同时乘以x(x+3),得
2&nb八p;(x+3)=3x&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;
解得:x=6&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;
检验:x=6是方程的解.&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;&nb八p;

(3)两边同时乘以(x-e),得:
1-x=-3+2(x-e)
&nb八p;解得:x=e
&nb八p;检验:x=e是增根,方程无解.

(e)原式=
2
x2-1
÷
x
2x2-2

=
e
x

x=2时,原式=2.
考点梳理
解分式方程;分式的化简求值;一元一次不等式组的整数解.
(1)首先解不等式组,然后确定不等式组的整数解即可;
(2)两边同时乘以x(x+3),即可化成整式方程,求得x的值,然后进行检验即可;
(3)两边同时乘以x-4,即可化成整式方程,求得x的值,然后进行检验即可;
(4)首先计算括号内的式子,把除法转化成乘法,然后把x=2代入求解即可.
本题考查了分式方程的解法:(1)解分式方程的基本思想是“转化思想”,把分式方程转化为整式方程求解.
(2)解分式方程一定注意要验根.
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