题目内容
计算
①(2.25÷0.375×
-
×2
)÷1
②[(9
-
)÷2
]×[1÷(2
-2
)]
③1994
×79+790×
+7.9×31
④1+2-3-4+5+6-7-8+…+2001+2002.
①(2.25÷0.375×
| 1 |
| 6 |
| 3 |
| 10 |
| 2 |
| 3 |
| 1 |
| 4 |
②[(9
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 10 |
| 9 |
| 100 |
③1994
| 1 |
| 2 |
| 6 |
| 250 |
④1+2-3-4+5+6-7-8+…+2001+2002.
考点:整数、分数、小数、百分数四则混合运算,加减法中的巧算
专题:运算顺序及法则,运算定律及简算
分析:(1)(2)根据四则混合运算的顺序进行计算即可.
(3)根据乘法分配律进行计算.
(4)每4项组合,例如1+2-3-4=5+6-7-8=-4,从1到2002共2002项,去除最后两项还有2000项,可分成:(2002-2)÷4=500(组),所以和为500×(-4)+2001+2002,计算即可.
(3)根据乘法分配律进行计算.
(4)每4项组合,例如1+2-3-4=5+6-7-8=-4,从1到2002共2002项,去除最后两项还有2000项,可分成:(2002-2)÷4=500(组),所以和为500×(-4)+2001+2002,计算即可.
解答:
解:①(2.25÷0.375×
-
×2
)÷1
=(2.25÷0.375×
-
×
)÷1
=(
÷
×
-
)÷1
=(
×
×
-
)÷1
=(6×
-
)÷1
=
×
=
②[(9
-
)÷2
]×[1÷(2
-2
)]
=[9×
]×[1÷(2
-2
)]
=3.6×[1÷
]
=3.6×100
=360
③1994
×79+790×
+7.9×31
=7.9×(19945+2.4+31)
=7.9×19978.4
=157829.36
④1+2-3-4+5+6-7-8+…+2001+2002
=(1+2-3-4)+(5+6-7-8)+…+(1997+1998-1999-2000)+2001+2002
=(-4)×500+2001+2002
=-2000+2001+2002
=2003
| 1 |
| 6 |
| 3 |
| 10 |
| 2 |
| 3 |
| 1 |
| 4 |
=(2.25÷0.375×
| 1 |
| 6 |
| 3 |
| 10 |
| 8 |
| 3 |
| 1 |
| 4 |
=(
| 9 |
| 4 |
| 3 |
| 8 |
| 1 |
| 6 |
| 4 |
| 5 |
| 1 |
| 4 |
=(
| 9 |
| 4 |
| 8 |
| 3 |
| 1 |
| 6 |
| 4 |
| 5 |
| 1 |
| 4 |
=(6×
| 1 |
| 6 |
| 4 |
| 5 |
| 1 |
| 4 |
=
| 1 |
| 5 |
| 4 |
| 5 |
=
| 4 |
| 25 |
②[(9
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 10 |
| 9 |
| 100 |
=[9×
| 2 |
| 5 |
| 10 |
| 100 |
| 9 |
| 100 |
=3.6×[1÷
| 1 |
| 100 |
=3.6×100
=360
③1994
| 1 |
| 2 |
| 6 |
| 250 |
=7.9×(19945+2.4+31)
=7.9×19978.4
=157829.36
④1+2-3-4+5+6-7-8+…+2001+2002
=(1+2-3-4)+(5+6-7-8)+…+(1997+1998-1999-2000)+2001+2002
=(-4)×500+2001+2002
=-2000+2001+2002
=2003
点评:本题考查的是分数的四则混合运算的知识,解答时根据数据特点灵活的进行计算.
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