题目内容
解方程.
|
|
x×(
|
分析:(1)根据等式的性质,两边同加上
,再同乘3即可;
(2)根据等式的性质,两边同减去
,再同乘
即可;
(3)原式变为
x=
,根据等式的性质,两边同乘
即可.
| 3 |
| 4 |
(2)根据等式的性质,两边同减去
| 3 |
| 5 |
| 5 |
| 2 |
(3)原式变为
| 11 |
| 15 |
| 22 |
| 19 |
| 15 |
| 11 |
解答:解:(1)
x-
=0,
x-
+
=0+
,
x=
,
x×3=
×3,
x=
;
(2)
+
x=1,
+
x-
=1-
,
x=
,
x×
=
×
,
x=1;
(3)x×(
+
)=
,
x=
,
x×
=
×
,
x=
.
| 1 |
| 3 |
| 3 |
| 4 |
| 1 |
| 3 |
| 3 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
| 1 |
| 3 |
| 3 |
| 4 |
| 1 |
| 3 |
| 3 |
| 4 |
x=
| 9 |
| 4 |
(2)
| 3 |
| 5 |
| 2 |
| 5 |
| 3 |
| 5 |
| 2 |
| 5 |
| 3 |
| 5 |
| 3 |
| 5 |
| 2 |
| 5 |
| 2 |
| 5 |
| 2 |
| 5 |
| 5 |
| 2 |
| 2 |
| 5 |
| 5 |
| 2 |
x=1;
(3)x×(
| 1 |
| 3 |
| 2 |
| 5 |
| 22 |
| 19 |
| 11 |
| 15 |
| 22 |
| 19 |
| 11 |
| 15 |
| 15 |
| 11 |
| 22 |
| 19 |
| 15 |
| 11 |
x=
| 30 |
| 19 |
点评:在解方程时应根据等式的性质,即等式两边同加上、同减去、同乘上或同除以某一个数(0除外),等式的两边仍相等,同时注意“=”上下要对齐.
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