题目内容
|
X:
|
|
X+
|
分析:(1)根据等式的性质,两边同乘
即可;
(2)根据比例的性质改写成X=
×
,计算即可;
(3)原式变为
X=
,根据等式的性质,两边同乘
即可;
(4)原式变为
X=
,根据等式的性质,两边同乘
即可.
| 8 |
| 5 |
(2)根据比例的性质改写成X=
| 4 |
| 7 |
| 2 |
| 3 |
(3)原式变为
| 2 |
| 5 |
| 1 |
| 6 |
| 5 |
| 2 |
(4)原式变为
| 16 |
| 9 |
| 4 |
| 3 |
| 9 |
| 16 |
解答:解:(1)
X=40,
X×
=40×
,
X=64;
(2)X:
=
,
X=
×
,
X=
;
(3)
X=
×
,
X=
,
X×
=
×
,
X=
;
(4)X+
X=
,
X=
,
X×
=
×
,
X=
.
| 5 |
| 8 |
| 5 |
| 8 |
| 8 |
| 5 |
| 8 |
| 5 |
X=64;
(2)X:
| 4 |
| 7 |
| 2 |
| 3 |
X=
| 4 |
| 7 |
| 2 |
| 3 |
X=
| 8 |
| 21 |
(3)
| 2 |
| 5 |
| 4 |
| 9 |
| 3 |
| 8 |
| 2 |
| 5 |
| 1 |
| 6 |
| 2 |
| 5 |
| 5 |
| 2 |
| 1 |
| 6 |
| 5 |
| 2 |
X=
| 5 |
| 12 |
(4)X+
| 7 |
| 9 |
| 4 |
| 3 |
| 16 |
| 9 |
| 4 |
| 3 |
| 16 |
| 9 |
| 9 |
| 16 |
| 4 |
| 3 |
| 9 |
| 16 |
X=
| 3 |
| 4 |
点评:此题考查了根据等式的性质解方程,即等式两边同加上、同减去、同乘上或同除以一个不为0的数,等式仍相等.同时注意“=”上下要对齐.
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