题目内容
| 解方程: Χ-
|
1-
|
(1+
|
分析:(1)原式变为
x=
,根据等式的性质,两边同乘
即可;
(2)根据等式的性质,两边同加上
x,得
+
x=1,两边同减去
,再同乘4即可;
(3)原式变为
x=
,根据等式的性质,两边同乘
即可.
| 3 |
| 4 |
| 9 |
| 32 |
| 4 |
| 3 |
(2)根据等式的性质,两边同加上
| 1 |
| 4 |
| 9 |
| 32 |
| 1 |
| 4 |
| 9 |
| 32 |
(3)原式变为
| 7 |
| 5 |
| 14 |
| 15 |
| 5 |
| 7 |
解答:解:(1)x-
x=
x=
x×
=
×
x=
;
(2)1-
x=
1-
x+
x=
+
x
+
x=1
+
x-
=1-
x=
x×4=
×4
x=
;
(3)(1+
)x=
x=
x×
=
×
x=
.
| 1 |
| 4 |
| 9 |
| 32 |
| 3 |
| 4 |
| 9 |
| 32 |
| 3 |
| 4 |
| 4 |
| 3 |
| 9 |
| 32 |
| 4 |
| 3 |
x=
| 3 |
| 8 |
(2)1-
| 1 |
| 4 |
| 9 |
| 32 |
1-
| 1 |
| 4 |
| 1 |
| 4 |
| 9 |
| 32 |
| 1 |
| 4 |
| 9 |
| 32 |
| 1 |
| 4 |
| 9 |
| 32 |
| 1 |
| 4 |
| 9 |
| 32 |
| 9 |
| 32 |
| 1 |
| 4 |
| 23 |
| 32 |
| 1 |
| 4 |
| 23 |
| 32 |
x=
| 23 |
| 8 |
(3)(1+
| 2 |
| 5 |
| 14 |
| 15 |
| 7 |
| 5 |
| 14 |
| 15 |
| 7 |
| 5 |
| 5 |
| 7 |
| 14 |
| 15 |
| 5 |
| 7 |
x=
| 2 |
| 3 |
点评:此题考查了运用等式的性质解方程,即等式两边同加上或同减去、同乘上或同除以一个数(0除外),两边仍相等,同时注意“=”上下要对齐.
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