题目内容
（2009?中山市）脱式计算．
①0.9÷[0.69÷(
+0.9)-(
÷0.45-1
)]
②[(3.7+5.4×4
)×(1÷
-1.0625)]+(10
+2
)-1.9
③[5
-(4-3
)+(3
-
)]×[7.6÷
+(3.6-1.2)×1.25]
④[2.5-(4
-2
)]÷[(8-7.2)×
÷2
]
⑤[4
-1
×(0.25+
)]÷(0.75+4
)．
①0.9÷[0.69÷(
| 1 |
| 4 |
| 3 |
| 4 |
| 2 |
| 5 |
②[(3.7+5.4×4
| 11 |
| 13 |
| 16 |
| 17 |
| 4 |
| 7 |
| 3 |
| 7 |
③[5
| 5 |
| 9 |
| 1 |
| 5 |
| 7 |
| 9 |
| 4 |
| 3 |
| 4 |
| 5 |
④[2.5-(4
| 1 |
| 3 |
| 5 |
| 6 |
| 2 |
| 3 |
| 2 |
| 3 |
⑤[4
| 14 |
| 15 |
| 4 |
| 5 |
| 5 |
| 12 |
| 7 |
| 12 |
分析：（1）（2）按照先算小括号里面的第二级运算，再算小括号里面的第一级运算，然后算中括号里面，最后算括号外面的顺序解答，
（3）（5）按照先算小括号里面的，再算中括号里面的第二级运算，最后按照从左到右顺序算中括号里面的顺序解答，
（4）按照先算小括号里面的，再算中括号里面的顺序解答．
（3）（5）按照先算小括号里面的，再算中括号里面的第二级运算，最后按照从左到右顺序算中括号里面的顺序解答，
（4）按照先算小括号里面的，再算中括号里面的顺序解答．
解答：解：①0.9÷[0.69÷(
+0.9)-(
÷0.45-1
)]
=0.9÷[0.69÷
-（
-1
）]
=0.9÷[0.69÷
-
]
=0.9÷[0.6-
]
=0.9÷
=2.7；
②[(3.7+5.4×4
)×(1÷
-1.0625)]+(10
+2
)-1.9
=[（3.7+26
）]×（1
-1.0625）+13-1.9
=29
×0+13-1.9
=13-1.9
=11.1；
③[5
-(4-3
)+(3
-
)]×[7.6÷
+(3.6-1.2)×1.25]
=[5
-
+2
]×[9.5+2.4×1.25]
=7
×[9.5+3]
=7
×12.5
=90；
④[2.5-(4
-2
)]÷[(8-7.2)×
÷2
]
=[2.5-1.5]÷[0.8×
÷2
]
=1÷[
÷2
]
=1÷
=5；
⑤[4
-1
×(0.25+
)]÷(0.75+4
)
=[4
-1
×
]÷5
=[4
-1
]÷5
=3
÷5
=
．
| 1 |
| 4 |
| 3 |
| 4 |
| 2 |
| 5 |
=0.9÷[0.69÷
| 23 |
| 20 |
| 5 |
| 3 |
| 2 |
| 5 |
=0.9÷[0.69÷
| 23 |
| 20 |
| 4 |
| 15 |
=0.9÷[0.6-
| 4 |
| 15 |
=0.9÷
| 1 |
| 3 |
=2.7；
②[(3.7+5.4×4
| 11 |
| 13 |
| 16 |
| 17 |
| 4 |
| 7 |
| 3 |
| 7 |
=[（3.7+26
| 11 |
| 65 |
| 1 |
| 16 |
=29
| 113 |
| 130 |
=13-1.9
=11.1；
③[5
| 5 |
| 9 |
| 1 |
| 5 |
| 7 |
| 9 |
| 4 |
| 3 |
| 4 |
| 5 |
=[5
| 5 |
| 9 |
| 4 |
| 5 |
| 4 |
| 9 |
=7
| 1 |
| 5 |
=7
| 1 |
| 5 |
=90；
④[2.5-(4
| 1 |
| 3 |
| 5 |
| 6 |
| 2 |
| 3 |
| 2 |
| 3 |
=[2.5-1.5]÷[0.8×
| 2 |
| 3 |
| 2 |
| 3 |
=1÷[
| 8 |
| 15 |
| 2 |
| 3 |
=1÷
| 1 |
| 5 |
=5；
⑤[4
| 14 |
| 15 |
| 4 |
| 5 |
| 5 |
| 12 |
| 7 |
| 12 |
=[4
| 14 |
| 15 |
| 4 |
| 5 |
| 2 |
| 3 |
| 1 |
| 3 |
=[4
| 14 |
| 15 |
| 1 |
| 5 |
| 1 |
| 3 |
=3
| 11 |
| 15 |
| 1 |
| 3 |
=
| 7 |
| 10 |
点评：本题主要考查学生依据四则运算计算方法正确进行计算的能力，关键是计算结果要准确．
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