题目内容
(2012?中山模拟)9.36÷[(6-2.88)×(5
-1.875)]
(2
×
+2
×6.2-5.8×2
-
÷
)×
(0.1+
+7
÷1.25)÷(18.25-5
×3
)
(96÷4-4
×5)×[(3
-3.875)×49.5÷0.8÷
].
| 7 |
| 8 |
(2
| 2 |
| 9 |
| 4 |
| 5 |
| 2 |
| 9 |
| 2 |
| 9 |
| 1 |
| 5 |
| 9 |
| 20 |
| 9 |
| 20 |
(0.1+
| 2 |
| 5 |
| 1 |
| 8 |
| 1 |
| 2 |
| 1 |
| 5 |
(96÷4-4
| 1 |
| 18 |
| 7 |
| 8 |
| 4 |
| 5 |
分析:算式(1)(3)(4)根据四则混合运算的运算顺序计算即可:先算乘除,再算加减,有括号的要先算括号里面的;
算式(2)可根据乘法分配律计算.
算式(2)可根据乘法分配律计算.
解答:解:(1)9.36÷[(6-2.88)×(5
-1.875)]
=9.36÷[3.12×(5.875-1.875)],
=9.36÷[3.12×4],
=9.36÷12.48,
=0.75;
(2)(2
×
+2
×6.2-5.8×2
-
÷
)×
=(
×0.8+
×6.2-5.8×
-0.2×
)×
,
=(0.8+6.2-5.8-0.2)×
×
,
=1×1,
=1;
(3)(0.1+
+7
÷1.25)÷(18.25-5
×3
)
=(0.1+0.4+7.125÷1.25)÷(18.25-5.5×3.2),
=(0.5+5.7)÷(18.25-17.6),
=6.2÷0.65,
=9
;
(4)(96÷4-4
×5)×[(3
-3.875)×49.5÷0.8÷
]
=(24-20
)×[(3.875-3.875)×49.5÷0.8÷
],
=3
×[0×49.5÷0.8÷
],
=3
×0,
=0.
| 7 |
| 8 |
=9.36÷[3.12×(5.875-1.875)],
=9.36÷[3.12×4],
=9.36÷12.48,
=0.75;
(2)(2
| 2 |
| 9 |
| 4 |
| 5 |
| 2 |
| 9 |
| 2 |
| 9 |
| 1 |
| 5 |
| 9 |
| 20 |
| 9 |
| 20 |
=(
| 20 |
| 9 |
| 20 |
| 9 |
| 20 |
| 9 |
| 20 |
| 9 |
| 9 |
| 20 |
=(0.8+6.2-5.8-0.2)×
| 20 |
| 9 |
| 9 |
| 20 |
=1×1,
=1;
(3)(0.1+
| 2 |
| 5 |
| 1 |
| 8 |
| 1 |
| 2 |
| 1 |
| 5 |
=(0.1+0.4+7.125÷1.25)÷(18.25-5.5×3.2),
=(0.5+5.7)÷(18.25-17.6),
=6.2÷0.65,
=9
| 7 |
| 13 |
(4)(96÷4-4
| 1 |
| 18 |
| 7 |
| 8 |
| 4 |
| 5 |
=(24-20
| 5 |
| 18 |
| 4 |
| 5 |
=3
| 13 |
| 18 |
| 4 |
| 5 |
=3
| 13 |
| 18 |
=0.
点评:当算式中同时含有分数与小数时,要根据式中数据的特点灵活将它们互化为再进行计算.
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