题目内容
计算:
(1)3
×(-
)-(-
)×2
-
×(-
);
(2)2
-(-8
)+(-2
)+0.25-1.5-2.75;
(3)
÷(-2
)-
×(-1
)-0.75;
(4)﹛4
+[-9×(+
)-0.8]﹜÷(-5
).
(1)3
| 1 |
| 2 |
| 5 |
| 7 |
| 5 |
| 7 |
| 1 |
| 2 |
| 5 |
| 7 |
| 1 |
| 2 |
(2)2
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 4 |
(3)
| 2 |
| 5 |
| 2 |
| 5 |
| 8 |
| 21 |
| 3 |
| 4 |
(4)﹛4
| 1 |
| 2 |
| 1 |
| 9 |
| 2 |
| 5 |
分析:(1)先变形得到原式=
×(-
)+
×
+
×
,然后根据乘法的分配律进行计算;
(2)先把小数化为分数,再利用加法的交换律得到原式=2
-2
+
-2
+8
-1
,然后进行同分母的加减运算;
(3)先把小数化为分数、带分数化为假分数和除法化为乘法得到原式=
×(-
)+
×
-
,然后约分后进行加减运算;
(4)先带分数化为假分数除法化为乘法,再算括号内的乘法运算,然后根据乘法的分配律进行计算.
| 7 |
| 2 |
| 5 |
| 7 |
| 5 |
| 7 |
| 5 |
| 2 |
| 5 |
| 7 |
| 1 |
| 2 |
(2)先把小数化为分数,再利用加法的交换律得到原式=2
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 4 |
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
(3)先把小数化为分数、带分数化为假分数和除法化为乘法得到原式=
| 2 |
| 5 |
| 5 |
| 12 |
| 8 |
| 21 |
| 7 |
| 4 |
| 3 |
| 4 |
(4)先带分数化为假分数除法化为乘法,再算括号内的乘法运算,然后根据乘法的分配律进行计算.
解答:解:(1)原式=
×(-
)+
×
+
×
=
×(-
+
+
)
=
×(-
)
=-
;
(2)原式=2
+8
-2
+
-1
-2
=2
-2
+
-2
+8
-1
=-2+7
=5;
(3)原式=
×(-
)+
×
-
=-
+
-
=-
+
-
=-
;
(4)原式={
-1-
}×(-
)
=
×(-
)-1×(-
)+
×
=-
+
+
=-
+
=-
.
| 7 |
| 2 |
| 5 |
| 7 |
| 5 |
| 7 |
| 5 |
| 2 |
| 5 |
| 7 |
| 1 |
| 2 |
=
| 5 |
| 7 |
| 7 |
| 2 |
| 5 |
| 2 |
| 1 |
| 2 |
=
| 5 |
| 7 |
| 1 |
| 2 |
=-
| 5 |
| 14 |
(2)原式=2
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 4 |
=2
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 4 |
| 3 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
=-2+7
=5;
(3)原式=
| 2 |
| 5 |
| 5 |
| 12 |
| 8 |
| 21 |
| 7 |
| 4 |
| 3 |
| 4 |
=-
| 1 |
| 6 |
| 2 |
| 3 |
| 3 |
| 4 |
=-
| 2 |
| 12 |
| 8 |
| 12 |
| 9 |
| 12 |
=-
| 1 |
| 4 |
(4)原式={
| 9 |
| 2 |
| 4 |
| 5 |
| 5 |
| 27 |
=
| 9 |
| 2 |
| 5 |
| 27 |
| 5 |
| 27 |
| 4 |
| 5 |
| 5 |
| 27 |
=-
| 5 |
| 6 |
| 5 |
| 27 |
| 4 |
| 27 |
=-
| 5 |
| 6 |
| 2 |
| 6 |
=-
| 1 |
| 2 |
点评:本题考查了有理数的混合运算:先算乘方,再算乘除,然后进行加减运算;有括号先算括号.
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