题目内容
用递等式计算(能简算的要简算)
①(
+
)÷(1-
×
)
②3.2×1.25+4.8×1
③[(5
-4
)÷1
]×25%
④
+(
-
)÷
.
①(
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 8 |
| 3 |
| 4 |
②3.2×1.25+4.8×1
| 1 |
| 4 |
③[(5
| 2 |
| 3 |
| 3 |
| 4 |
| 5 |
| 6 |
④
| 15 |
| 16 |
| 7 |
| 16 |
| 1 |
| 4 |
| 1 |
| 2 |
分析:1)把
+
和
×
同时计算,再做1-
,再算括号外面的除;
(2)把1
看成1.25,利用乘法分配律,据此进行计算即可;
(3)先算小括号里,再算中括号里,最后算括号外面的;
(4)(
-
)÷
利用乘法分配律计算,最后算加.
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 8 |
| 3 |
| 4 |
| 9 |
| 32 |
(2)把1
| 1 |
| 4 |
(3)先算小括号里,再算中括号里,最后算括号外面的;
(4)(
| 7 |
| 16 |
| 1 |
| 4 |
| 1 |
| 2 |
解答:①(
+
)÷(1-
×
);
=
÷(1-
),
=
×
,
=1
;
②3.2×1.25+4.8×1
=3.2×1.25+4.8×1.25,
=1.25×(3.2+4.8),
=1.25×8,
=10;
③[(5
-4
)÷1
]×25%,
=[
÷
]×
,
=[
×
]×
,
=
×
,
=
;
④
+(
-
)÷
,
=
+
÷
,
=
+
×2,
=
+
,
=1
.
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 8 |
| 3 |
| 4 |
=
| 3 |
| 4 |
| 9 |
| 32 |
=
| 3 |
| 4 |
| 32 |
| 23 |
=1
| 1 |
| 23 |
②3.2×1.25+4.8×1
| 1 |
| 4 |
=3.2×1.25+4.8×1.25,
=1.25×(3.2+4.8),
=1.25×8,
=10;
③[(5
| 2 |
| 3 |
| 3 |
| 4 |
| 5 |
| 6 |
=[
| 11 |
| 12 |
| 11 |
| 6 |
| 1 |
| 4 |
=[
| 11 |
| 12 |
| 6 |
| 11 |
| 1 |
| 4 |
=
| 1 |
| 2 |
| 1 |
| 4 |
=
| 1 |
| 8 |
④
| 15 |
| 16 |
| 7 |
| 16 |
| 1 |
| 4 |
| 1 |
| 2 |
=
| 15 |
| 16 |
| 3 |
| 16 |
| 1 |
| 2 |
=
| 15 |
| 16 |
| 3 |
| 16 |
=
| 15 |
| 16 |
| 3 |
| 8 |
=1
| 5 |
| 16 |
点评:此题考查整数、分数、小数、百分数的四则混合运算,能简算要简算;不能简算的,有括号时,要按照先算括号里面的,再算括号外面的;没有括号时,要按照先算乘除法,再算加减法.
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