题目内容
解方程
9x-3=
x÷
+
=
x+
x=
÷x×
=
.
9x-3=
| 3 |
| 8 |
x÷
| 5 |
| 6 |
| 1 |
| 5 |
| 4 |
| 5 |
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 2 |
| 3 |
| 7 |
| 1 |
| 3 |
| 5 |
| 14 |
分析:(1)根据等式性质,两边同加3,再同除以9即可;
(2)根据等式性质,两边同减去
,再同乘
即可;
(3)先运用乘法分配律的逆运算,把原式变为(
+
)x=
,即
x=
,再根据等式性质,两边同除以
即可;
(4)根据等式性质,两边同乘3,变为
÷x=
,再同乘x,变为
×x=
,两边再同除以
,计算得解.
(2)根据等式性质,两边同减去
| 1 |
| 5 |
| 5 |
| 6 |
(3)先运用乘法分配律的逆运算,把原式变为(
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
(4)根据等式性质,两边同乘3,变为
| 3 |
| 7 |
| 15 |
| 14 |
| 15 |
| 14 |
| 3 |
| 7 |
| 15 |
| 14 |
解答:解:(1)9x-3=
,
9x-3+3=
+3,
9x=
,
9x÷9=
÷9,
x=
×
,
x=
;
(2)x÷
+
=
,
x÷
+
-
=
-
,
x÷
=
,
x÷
×
=
×
,
x=
;
(3)
x+
x=
,
(
+
)x=
,
x=
,
x÷
=
÷
,
x=
×
,
x=
;
(4)
÷x×
=
,
÷x×
×3=
×3,
÷x=
,
÷x×x=
×x,
×x=
,
×x÷
=
÷
,
x=
×
,
x=
.
| 3 |
| 8 |
9x-3+3=
| 3 |
| 8 |
9x=
| 27 |
| 8 |
9x÷9=
| 27 |
| 8 |
x=
| 27 |
| 8 |
| 1 |
| 9 |
x=
| 3 |
| 8 |
(2)x÷
| 5 |
| 6 |
| 1 |
| 5 |
| 4 |
| 5 |
x÷
| 5 |
| 6 |
| 1 |
| 5 |
| 1 |
| 5 |
| 4 |
| 5 |
| 1 |
| 5 |
x÷
| 5 |
| 6 |
| 3 |
| 5 |
x÷
| 5 |
| 6 |
| 5 |
| 6 |
| 3 |
| 5 |
| 5 |
| 6 |
x=
| 1 |
| 2 |
(3)
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 2 |
(
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
| 5 |
| 6 |
| 1 |
| 2 |
| 5 |
| 6 |
x=
| 1 |
| 2 |
| 6 |
| 5 |
x=
| 3 |
| 5 |
(4)
| 3 |
| 7 |
| 1 |
| 3 |
| 5 |
| 14 |
| 3 |
| 7 |
| 1 |
| 3 |
| 5 |
| 14 |
| 3 |
| 7 |
| 15 |
| 14 |
| 3 |
| 7 |
| 15 |
| 14 |
| 15 |
| 14 |
| 3 |
| 7 |
| 15 |
| 14 |
| 15 |
| 14 |
| 3 |
| 7 |
| 15 |
| 14 |
x=
| 3 |
| 7 |
| 14 |
| 15 |
x=
| 2 |
| 5 |
点评:解方程的依据是等式性质,同时注意等号对齐.
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