题目内容
解方程.
|
x+
|
4-
|
分析:(1)先根据比例的性质改写成
x=
,再根据等式的性质,两边同乘
即可;
(2)原式变为(1+
)x=
,即
x=
,根据等式的性质,两边同乘
即可;
(3)根据等式的性质,两边同加上
x,得
+
x=4,两边同减去
,再同乘
即可.
| 3 |
| 8 |
| 3 |
| 5 |
| 8 |
| 3 |
(2)原式变为(1+
| 3 |
| 7 |
| 3 |
| 7 |
| 10 |
| 7 |
| 3 |
| 7 |
| 7 |
| 10 |
(3)根据等式的性质,两边同加上
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 6 |
| 5 |
解答:解:(1)
:x=
,
x=
,
x×
=
×
,
x=
;
(2)x+
x=
,
(1+
)x=
,
x=
,
x×
=
×
,
x=
;
(3)4-
x=
,
4-
x+
x=
+
x,
+
x=4,
+
x-
=4-
,
x=
,
x×
=
×
,
x=
.
| 3 |
| 5 |
| 3 |
| 8 |
| 3 |
| 8 |
| 3 |
| 5 |
| 3 |
| 8 |
| 8 |
| 3 |
| 3 |
| 5 |
| 8 |
| 3 |
x=
| 8 |
| 5 |
(2)x+
| 3 |
| 7 |
| 3 |
| 7 |
(1+
| 3 |
| 7 |
| 3 |
| 7 |
| 10 |
| 7 |
| 3 |
| 7 |
| 10 |
| 7 |
| 7 |
| 10 |
| 3 |
| 7 |
| 7 |
| 10 |
x=
| 3 |
| 10 |
(3)4-
| 5 |
| 6 |
| 1 |
| 2 |
4-
| 5 |
| 6 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 1 |
| 2 |
| 5 |
| 6 |
| 7 |
| 2 |
| 5 |
| 6 |
| 6 |
| 5 |
| 7 |
| 2 |
| 6 |
| 5 |
x=
| 21 |
| 5 |
点评:此题考查了根据等式的性质解方程,即等式两边同加上或同减去、同乘上或同除以同一个数(0除外),两边仍相等.
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