题目内容
解方程或比例.
|
|
分析:(1)根据等式的性质,两边同时加上
x-
,可得
=
x,等式的两边交换位置,等式仍然成立可得:
x=
,再利用等式的性质两边同时乘5即可;
(2)根据比例的基本性质可得:
x=
×8,再利用等式的性质,两边同时乘
,即可解答.
| 1 |
| 5 |
| 1 |
| 10 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 2 |
(2)根据比例的基本性质可得:
| 4 |
| 9 |
| 2 |
| 3 |
| 9 |
| 4 |
解答:解:
-
x=
,
-
x+
x-
=
+
x-
,
=
x,
x=
,
x×5=
×5,
x=
;
:
=8:x.
x=
×8,
x×
=
×8×
,
x=12.
| 3 |
| 5 |
| 1 |
| 5 |
| 1 |
| 10 |
| 3 |
| 5 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 10 |
| 1 |
| 10 |
| 1 |
| 5 |
| 1 |
| 10 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 2 |
x=
| 5 |
| 2 |
| 4 |
| 9 |
| 2 |
| 3 |
| 4 |
| 9 |
| 2 |
| 3 |
| 4 |
| 9 |
| 9 |
| 4 |
| 2 |
| 3 |
| 9 |
| 4 |
x=12.
点评:此题主要考查等式的性质和比例的基本性质的灵活应用.
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