题目内容
| 递等式计算: ( | 1÷(55-5×7 )× |
| ( | 128× |
解:
(1)(
-
)×
×36,
=(
-
)×6,
=
×6-
×6,
=3-1,
=2;
(2)1÷(55-5×7)×
,
=1÷(55-35)×
,
=1÷20×
,
=
×
,
=
;
(3)(
×
+
)÷(11-
),
=(
+
)÷9
,
=
÷9
,
=
;
(4)128×
-
×128-2÷
,
=128×4.5-0.5×128-2×128,
=128×(4.5-0.5-2),
=128×2,
=256.
分析:(1)根据乘法分配律进行计算;
(2)先算小括号里面的乘法,再算小括号里面的减法,再算除法,最后算括号外面的乘法;
(3)先算小括号里面的乘法和减法,再算小括号里面的加法,最后算除法;
(4)根据乘法分配律进行计算.
点评:四则运算,先弄清运算顺序,然后再进一步计算,能简算的要简算.
(1)(
=(
=
=3-1,
=2;
(2)1÷(55-5×7)×
=1÷(55-35)×
=1÷20×
=
=
(3)(
=(
=
=
(4)128×
=128×4.5-0.5×128-2×128,
=128×(4.5-0.5-2),
=128×2,
=256.
分析:(1)根据乘法分配律进行计算;
(2)先算小括号里面的乘法,再算小括号里面的减法,再算除法,最后算括号外面的乘法;
(3)先算小括号里面的乘法和减法,再算小括号里面的加法,最后算除法;
(4)根据乘法分配律进行计算.
点评:四则运算,先弄清运算顺序,然后再进一步计算,能简算的要简算.
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