题目内容
(1)-2
-(+3
)+2.25-(-
)
(2)2
×|
-
|×
÷(-1
)
(3)0.5×19
-2
×13+0.5×
+
×(-13)
(4)-24×(-
)3÷(-2)2+(-22)÷(-
)
(5)
am-4am-1+
am+4am-1-3(m为整数且m>1)
| 1 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
(2)2
| 1 |
| 5 |
| 1 |
| 3 |
| 1 |
| 2 |
| 3 |
| 11 |
| 1 |
| 4 |
(3)0.5×19
| 4 |
| 9 |
| 3 |
| 4 |
| 5 |
| 9 |
| 1 |
| 4 |
(4)-24×(-
| 3 |
| 2 |
| 23 |
| 3 |
(5)
| 3 |
| 4 |
| 1 |
| 2 |
分析:(1)先去括号,再把小数化为分数,然后将分母相同的分数相加,算出结果;
(2)先算绝对值,然后将除法转化为乘法计算;
(3)逆用分配律进行计算;
(4)先算乘方,然后将除法转化为乘法进行计算;
(5)合并同类项即可.
(2)先算绝对值,然后将除法转化为乘法计算;
(3)逆用分配律进行计算;
(4)先算乘方,然后将除法转化为乘法进行计算;
(5)合并同类项即可.
解答:解:(1)原式=-2
-3
+2
+
=-3
+
=-3;
(2)原式=
×
×
×(-
)
=-
;
(3)原式=0.5×(19
+
)-13×(2
+
)
=0.5×20-13×3
=10-39
=-29;
(4)原式=-16×(-
)×
+(-4)×(-
)
=54×
+
=15;
(5)原式=(
+
)am-3
=
am-3.
| 1 |
| 4 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
=-3
| 1 |
| 2 |
| 1 |
| 2 |
=-3;
(2)原式=
| 11 |
| 5 |
| 1 |
| 6 |
| 3 |
| 11 |
| 4 |
| 5 |
=-
| 2 |
| 25 |
(3)原式=0.5×(19
| 4 |
| 9 |
| 5 |
| 9 |
| 3 |
| 4 |
| 1 |
| 4 |
=0.5×20-13×3
=10-39
=-29;
(4)原式=-16×(-
| 27 |
| 8 |
| 1 |
| 4 |
| 3 |
| 8 |
=54×
| 1 |
| 4 |
| 3 |
| 2 |
=15;
(5)原式=(
| 3 |
| 4 |
| 1 |
| 2 |
=
| 5 |
| 4 |
点评:本题考查了整式的加减、有理数的混合运算,熟悉各种运算法则是解题的关键.
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