题目内容
解方程:
(1)(
|
(2)(1-25%)x=36 | (3)30%x=90 | (4)x+20%x=40. |
分析:(1)先化简方程,再依据等式的性质,方程两边同时除以
求解,
(2)先化简方程,再依据等式的性质,方程两边同时除以75%求解,
(3)依据等式的性质,方程两边同时除以30%求解,
(4)先化简方程,再依据等式的性质,方程两边同时除以120%求解.
35 |
24 |
(2)先化简方程,再依据等式的性质,方程两边同时除以75%求解,
(3)依据等式的性质,方程两边同时除以30%求解,
(4)先化简方程,再依据等式的性质,方程两边同时除以120%求解.
解答:解:(1)(
+
)x=
,
x=
,
x÷
=
÷
,
x=
;
(2)(1-25%)x=36,
75%x=36,
75%x÷75%=36÷75%,
x=48;
(3)30%x=90,
30%x÷30%=90÷30%,
x=300;
(4)x+20%x=40,
120%x=40,
120%x÷120%=40÷120%,
x=3
.
5 |
6 |
5 |
8 |
7 |
20 |
35 |
24 |
7 |
20 |
35 |
24 |
35 |
24 |
7 |
20 |
35 |
24 |
x=
6 |
25 |
(2)(1-25%)x=36,
75%x=36,
75%x÷75%=36÷75%,
x=48;
(3)30%x=90,
30%x÷30%=90÷30%,
x=300;
(4)x+20%x=40,
120%x=40,
120%x÷120%=40÷120%,
x=3
1 |
3 |
点评:等式的性质是解方程的依据,解方程时要注意:(1)等号要对齐,(2)方程能化简先化简.
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