题目内容
计算或化简:
(1)
-
.
(2)
-
÷
(3)先化简 (1+
)÷
,然后请你给a选取一个合适的值,再求此时原式的值.
(1)
1 |
2m-6 |
3 |
m2-9 |
(2)
1 |
x+2 |
x2+2x+1 |
x+2 |
x2-1 |
x-1 |
(3)先化简 (1+
3 |
a-2 |
a+1 |
a2-4 |
分析:(1)利用分式基本性质,首先通分,再合并即可得出答案;
(2)首先将原式中分式因式分解,再利用乘以一个数等于乘以这个数的倒数,化简得出,再求解即可.
(3)这道求代数式值的题目,通常做法是先把代数式化简,然后再代入求值.
(2)首先将原式中分式因式分解,再利用乘以一个数等于乘以这个数的倒数,化简得出,再求解即可.
(3)这道求代数式值的题目,通常做法是先把代数式化简,然后再代入求值.
解答:解:(1)
-
.
=
-
,
=
-
,
=
,
=
;
(2)
-
÷
,
=
-
×
,
=
-
,
=
;
(3)原式=
×
,
=
×
,
=a+2,
取a=1,则原式=1+2=3.
1 |
2m-6 |
3 |
m2-9 |
=
1 |
2(m-3) |
3 |
(m-3)(m+3) |
=
(m+3) |
2(m-3)(m+3) |
6 |
2(m-3)(m+3) |
=
m-3 |
2(m-3)(m+3) |
=
1 |
2(m+3) |
(2)
1 |
x+2 |
x2+2x+1 |
x+2 |
x2-1 |
x-1 |
=
1 |
x+2 |
(x+1) 2 |
x+2 |
x-1 |
(x+1)(x-1) |
=
1 |
x+2 |
x+1 |
x+2 |
=
-x |
x+2 |
(3)原式=
a-2+3 |
a-2 |
a2-4 |
a+1 |
=
a+1 |
a-2 |
(a+2)(a-2) |
a+1 |
=a+2,
取a=1,则原式=1+2=3.
点评:此题考查了分式的化简求值,解答此题的关键是正确将原式中式子因式分解,再进行混合运算求出即可.
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