题目内容
(13.5-8
-4.75)÷[5
×(x+1
)÷1
]=
,则x= .
| 1 |
| 4 |
| 1 |
| 2 |
| 13 |
| 14 |
| 4 |
| 7 |
| 1 |
| 30 |
考点:整数、分数、小数、百分数四则混合运算
专题:运算顺序及法则
分析:先化简方程,根据等式的性质,两边同时乘以[
x+
],再两边同时减去
,然后两边同时除以
求解即可.
| 7 |
| 2 |
| 27 |
| 4 |
| 9 |
| 40 |
| 7 |
| 60 |
解答:
解:(13.5-8
-4.75)÷[5
×(x+1
)÷1
]=
[13.5-(8
+4.75)]÷[5
×(x+1
)÷1
]=
[(13.5-13]÷[5
×(x+1
)÷1
]=
0.5÷[
×(x+1
)]=
0.5÷[
x+
]=
0.5÷[
x+
]×[
x+
]=
×[
x+
]
0.5=
x+
0.5-
=
x+
-
0.275=
x
0.275÷
=
x÷
x=
;
故答案为:
.
| 1 |
| 4 |
| 1 |
| 2 |
| 13 |
| 14 |
| 4 |
| 7 |
| 1 |
| 30 |
[13.5-(8
| 1 |
| 4 |
| 1 |
| 2 |
| 13 |
| 14 |
| 4 |
| 7 |
| 1 |
| 30 |
[(13.5-13]÷[5
| 1 |
| 2 |
| 13 |
| 14 |
| 4 |
| 7 |
| 1 |
| 30 |
0.5÷[
| 7 |
| 2 |
| 13 |
| 14 |
| 1 |
| 30 |
0.5÷[
| 7 |
| 2 |
| 27 |
| 4 |
| 1 |
| 30 |
0.5÷[
| 7 |
| 2 |
| 27 |
| 4 |
| 7 |
| 2 |
| 27 |
| 4 |
| 1 |
| 30 |
| 7 |
| 2 |
| 27 |
| 4 |
0.5=
| 7 |
| 60 |
| 9 |
| 40 |
0.5-
| 9 |
| 40 |
| 7 |
| 60 |
| 9 |
| 40 |
| 9 |
| 40 |
0.275=
| 7 |
| 60 |
0.275÷
| 7 |
| 60 |
| 7 |
| 60 |
| 7 |
| 60 |
x=
| 33 |
| 14 |
故答案为:
| 33 |
| 14 |
点评:此题考查的目的是理解方程的意义,掌握利用等式的性质解方程的方法步骤.
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