题目内容
1÷2x=
-
x=
x÷
=
x-
x=15
x×20%=20
x÷
×
=
.
1 |
2 |
5 |
7 |
10 |
21 |
5 |
14 |
6 |
5 |
3 |
7 |
7 |
3 |
x-
5 |
8 |
x×20%=20
x÷
6 |
7 |
9 |
14 |
3 |
4 |
分析:(1)根据等式的性质,两边同乘2x,得x=1;
(2)根据等式的性质,两边同加上
x,得
x+
=
,两边同减去
,再同除以
即可;
(3)根据等式的性质,两边同乘
,再同除以
即可;
(4)先运用乘法分配律的逆运算,把原式变为(1-
)x=15,即
x=15,再根据等式的性质,两边同除以
和即可;
(5)把20%化成小数0.2,根据等式的性质,两边同除以0.2和即可;
(6)先把除法改为乘法,求出
×
=
,原式变为x×
=
,再根据等式的性质,两边同除以
即可.
(2)根据等式的性质,两边同加上
10 |
21 |
10 |
21 |
5 |
14 |
5 |
7 |
15 |
14 |
10 |
21 |
(3)根据等式的性质,两边同乘
3 |
7 |
6 |
5 |
(4)先运用乘法分配律的逆运算,把原式变为(1-
5 |
8 |
3 |
8 |
3 |
8 |
(5)把20%化成小数0.2,根据等式的性质,两边同除以0.2和即可;
(6)先把除法改为乘法,求出
7 |
6 |
9 |
14 |
3 |
4 |
3 |
4 |
3 |
4 |
3 |
4 |
解答:解:(1)1÷2x=
,
1÷2x×2x=
×2x,
x=1;
(2)
-
x=
,
-
x+
x=
+
x,
x+
=
,
x+
-
=
-
,
x=
,
x÷
=
÷
,
x=
×
,
x=
;
(3)
x÷
=
,
x÷
×
=
×
,
x=1,
x÷
=1÷
,
x=
;
(4)x-
x=15,
(1-
)x=15,
x=15,
x÷
=15÷
,
x=15×
,
x=40;
(5)x×20%=20,
0.2x=20,
0.2x÷0.2=20÷0.2,
x=100;
(6)x÷
×
=
,
x×
×
=
,
x×
=
,
x×
÷
=
÷
,
x=1.
1 |
2 |
1÷2x×2x=
1 |
2 |
x=1;
(2)
5 |
7 |
10 |
21 |
5 |
14 |
5 |
7 |
10 |
21 |
10 |
21 |
5 |
14 |
10 |
21 |
10 |
21 |
5 |
14 |
5 |
7 |
10 |
21 |
5 |
14 |
5 |
14 |
5 |
7 |
5 |
14 |
10 |
21 |
5 |
14 |
10 |
21 |
10 |
21 |
5 |
14 |
10 |
21 |
x=
5 |
14 |
21 |
10 |
x=
3 |
4 |
(3)
6 |
5 |
3 |
7 |
7 |
3 |
6 |
5 |
3 |
7 |
3 |
7 |
7 |
3 |
3 |
7 |
6 |
5 |
6 |
5 |
6 |
5 |
6 |
5 |
x=
5 |
6 |
(4)x-
5 |
8 |
(1-
5 |
8 |
3 |
8 |
3 |
8 |
3 |
8 |
3 |
8 |
x=15×
8 |
3 |
x=40;
(5)x×20%=20,
0.2x=20,
0.2x÷0.2=20÷0.2,
x=100;
(6)x÷
6 |
7 |
9 |
14 |
3 |
4 |
x×
7 |
6 |
9 |
14 |
3 |
4 |
x×
3 |
4 |
3 |
4 |
x×
3 |
4 |
3 |
4 |
3 |
4 |
3 |
4 |
x=1.
点评:此题考查了根据等式的性质解方程,即方程两边同加、同减、同乘或同除以某数(0除外),方程的左右两边仍相等;注意“=”号上下要对齐.
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