题目内容
计算
(1)-40-28-(-19)+(-24)
(2)(1-1
-
+
)×(-24)
(3)-32-[22÷(-1)-13]×(-2)÷(-1)2008
(4)化简求值.2xy2-[5x-3(2x-1)-2xy2]+1,其中x=2,y=-
.
(1)-40-28-(-19)+(-24)
(2)(1-1
1 |
2 |
3 |
8 |
7 |
12 |
(3)-32-[22÷(-1)-13]×(-2)÷(-1)2008
(4)化简求值.2xy2-[5x-3(2x-1)-2xy2]+1,其中x=2,y=-
1 |
2 |
分析:(1)先把加法换成加法,再去掉加号得出-40-28+19-24,最后按加法法则计算即可;
(2)根据乘法的分配律展开得出1×(-24)-
×(-24)-
×(-24)+
×(-24),再算乘法,最后算加法即可;
(3)先算乘方,再算括号内的乘法(算乘法后算减法),最后算乘除法、减法即可;
(4)先去小括号、再去中括号,最后合并后代入求出即可.
(2)根据乘法的分配律展开得出1×(-24)-
3 |
2 |
3 |
8 |
7 |
12 |
(3)先算乘方,再算括号内的乘法(算乘法后算减法),最后算乘除法、减法即可;
(4)先去小括号、再去中括号,最后合并后代入求出即可.
解答:解:(1)-40-28-(-19)+(-24)
=-40-28+(+19)+(-24)
=-40-28+19-24
=-73;
(2)(1-1
-
+
)×(-24)
=1×(-24)-
×(-24)-
×(-24)+
×(-24)
=-24+36+9-14
-38+45
=7;
(3)原式=-9-[4÷(-1)-13]×(-2)÷1=-9-[-4-13]×(-2)÷1
=-9+17×(-2)÷1
=-9+(-34)
=-43;
(4)2xy2-[5x-3(2x-1)-2xy2]+1
=2xy2-[5x-6x+3-2xy2]+1
=2xy2-5x+6x-3+2xy2+1
=2xy2+x-2
当x=2,y=-
时,原式=2×2×(-
)2+2-2=2×2×
+2-2=1.
=-40-28+(+19)+(-24)
=-40-28+19-24
=-73;
(2)(1-1
1 |
2 |
3 |
8 |
7 |
12 |
=1×(-24)-
3 |
2 |
3 |
8 |
7 |
12 |
=-24+36+9-14
-38+45
=7;
(3)原式=-9-[4÷(-1)-13]×(-2)÷1=-9-[-4-13]×(-2)÷1
=-9+17×(-2)÷1
=-9+(-34)
=-43;
(4)2xy2-[5x-3(2x-1)-2xy2]+1
=2xy2-[5x-6x+3-2xy2]+1
=2xy2-5x+6x-3+2xy2+1
=2xy2+x-2
当x=2,y=-
1 |
2 |
1 |
2 |
1 |
4 |
点评:本题考查了有理数的混合运算和整式的混合运算,主要考查学生的计算能力,题目都比较好,但是一道比较容易出错的题目,一定要注意运算顺序.
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