题目内容
已知
=
,则
=
;已知
=
=
,则
=
.
a+b |
b |
5 |
2 |
a |
b |
2 |
3 |
2 |
3 |
2 |
a |
3 |
b |
4 |
c |
a+b+c |
a-b+c |
1 |
3 |
1 |
3 |
分析:(1)利用合比性质即可求解;
(2)设
=
=
=k,则a=
,b=
,c=
,代入代数式进行化简求值即可.
(2)设
2 |
a |
3 |
b |
4 |
c |
2 |
k |
3 |
k |
4 |
k |
解答:解:(1)
=
=
;
(2)设
=
=
=k,则a=
,b=
,c=
.
则
=
=
.
故答案是:
,
.
a |
b |
5-2 |
2 |
3 |
2 |
(2)设
2 |
a |
3 |
b |
4 |
c |
2 |
k |
3 |
k |
4 |
k |
则
a-b+c |
a+b+c |
| ||
|
1 |
3 |
故答案是:
3 |
2 |
1 |
3 |
点评:本题是基础题,考查了比例的基本性质,(2)中的设未知数的方法,是经常用到的.
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