题目内容
方程:
|
7×
|
40%Χ+30%Χ=21. |
分析:(1)根据等式的性质,两边同乘
即可;
(2)原式变为 1-
Χ=
,根据等式的性质,两边同加上
x,得
+
x=1,两边同减去
,再同乘
即可;
(3)先根据乘法分配律改写成(40%+30%)Χ=21,即0.7Χ=21,再根据等式的性质,两边同除以0.7即可.
| 6 |
| 5 |
(2)原式变为 1-
| 6 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 7 |
| 6 |
(3)先根据乘法分配律改写成(40%+30%)Χ=21,即0.7Χ=21,再根据等式的性质,两边同除以0.7即可.
解答:解:(1)
x=30,
x×
=30×
,
x=36;
(2)7×
-
Χ=
,
1-
Χ=
,
1-
Χ+
x=
+
x,
+
x=1,
+
x-
=1-
,
x=
,
x×
=
×
,
x=1;
(3)40%Χ+30%Χ=21,
(40%+30%)Χ=21,
0.7Χ=21,
0.7Χ÷0.7=21÷0.7,
Χ=30.
| 5 |
| 6 |
| 5 |
| 6 |
| 6 |
| 5 |
| 6 |
| 5 |
x=36;
(2)7×
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
1-
| 6 |
| 7 |
| 1 |
| 7 |
1-
| 6 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 1 |
| 7 |
| 1 |
| 7 |
| 6 |
| 7 |
| 6 |
| 7 |
| 6 |
| 7 |
| 7 |
| 6 |
| 6 |
| 7 |
| 7 |
| 6 |
x=1;
(3)40%Χ+30%Χ=21,
(40%+30%)Χ=21,
0.7Χ=21,
0.7Χ÷0.7=21÷0.7,
Χ=30.
点评:此题考查了根据等式的性质解方程,即等式两边同加、同减、同乘或同除以一个数(0除外),等式的左右两边仍相等;注意“=”上下要对齐.
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