题目内容
| 计算和解方程: (1)(0.375-
|
(2)
| ||||||||||||
(3)
|
(4)99-[1÷(
|
分析:(1)先算括号中的减法与加法,再算乘法.
(2)(3)根据等式的性质解方程即可:等式的两边同时乘或除以一个数,等式仍然成立.
(4)先算小括号中的减法,再算中括号中的除法,再算括号外的除法,最后算减法.
(2)(3)根据等式的性质解方程即可:等式的两边同时乘或除以一个数,等式仍然成立.
(4)先算小括号中的减法,再算中括号中的除法,再算括号外的除法,最后算减法.
解答:解:(1)(0.375-
)×(
+
)
=(
-
)×
=
×
=
;
(2)
x÷
=12
x÷
×
=12×
x÷
=3÷
x=
;
(3)
÷
x=
÷
x×
x=
×
x
÷
=
x÷
x=
;
(4)99-[1÷(
-
)]÷
=99-(1÷
)×
=99-
×
=99-9
=90.
| 1 |
| 8 |
| 3 |
| 4 |
| 1 |
| 3 |
=(
| 3 |
| 8 |
| 1 |
| 8 |
| 13 |
| 12 |
=
| 1 |
| 4 |
| 13 |
| 12 |
=
| 13 |
| 48 |
(2)
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 4 |
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
| 3 |
x=
| 9 |
| 2 |
(3)
| 7 |
| 12 |
| 2 |
| 3 |
| 29 |
| 10 |
| 7 |
| 12 |
| 2 |
| 3 |
| 2 |
| 3 |
| 29 |
| 10 |
| 2 |
| 3 |
| 7 |
| 12 |
| 29 |
| 15 |
| 29 |
| 15 |
| 29 |
| 15 |
x=
| 35 |
| 116 |
(4)99-[1÷(
| 2 |
| 3 |
| 1 |
| 4 |
| 4 |
| 15 |
=99-(1÷
| 5 |
| 12 |
| 15 |
| 4 |
=99-
| 12 |
| 5 |
| 15 |
| 4 |
=99-9
=90.
点评:此类题目由于不能运用简便方法计算,在计算时要细心,过程要完整.
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