题目内容
计算
(1)-
×[-32×(-
)2-2]÷(-1)2010
(2)(-
)×(-
)2+(-
)÷[(-
)3-
]
(3)(-8)÷2
+
×(-1)
(4)1-(
-
-
)×(-12)
(5)-(
-
+
-
+
)÷(-
)
(6)-32-(-4)2÷(-2)3+(-
)2010×(-2)2011+(-1)2011.
(1)-
| 3 |
| 4 |
| 2 |
| 3 |
(2)(-
| 9 |
| 5 |
| 5 |
| 3 |
| 3 |
| 8 |
| 1 |
| 2 |
| 1 |
| 4 |
(3)(-8)÷2
| 1 |
| 4 |
| 4 |
| 9 |
(4)1-(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 12 |
(5)-(
| 1 |
| 3 |
| 5 |
| 21 |
| 3 |
| 14 |
| 2 |
| 7 |
| 5 |
| 6 |
| 1 |
| 42 |
(6)-32-(-4)2÷(-2)3+(-
| 1 |
| 2 |
分析:(1)、(2)先算小括号里面的,再算中括号,乘方,乘法,除法,最后算加减即可;
(3)先算乘除,再算加减即可;
(4)先把括号中的每一项分别同-12相乘,再算加减;
(5)先把括号中的每一项分别同-
相除,再算加减;
(6)根据有理数混合运算的法则先算乘方,再算加减.
(3)先算乘除,再算加减即可;
(4)先把括号中的每一项分别同-12相乘,再算加减;
(5)先把括号中的每一项分别同-
| 1 |
| 42 |
(6)根据有理数混合运算的法则先算乘方,再算加减.
解答:解:(1)原式=-
×(-9×
-2)÷1
=-
×(-4-2)
=-
×(-6)
=
;
(2)原式=(-
)×
+(-
)÷[(-
)-
]
=-5+(-
)÷[-
-
]
=-5+(-
)×(-
)
=-5+1
=-4;
(3)原式=(-8)×
-
=-
-
=-4;
(4)原式=1-[
×(-12)-
×(-12)-
×(-12)]
=1-[-6+4+1]
=1+1
=2;
(5)原式=(
-
+
-
+
)×42
=
×42-
×42+
×42-
×42+
×42
=14-10+9-12+35
=36;
(6)原式=-9-16÷(-8)+[(-
)×(-2)]2011-1
=-9+2+1-1
=-7.
| 3 |
| 4 |
| 4 |
| 9 |
=-
| 3 |
| 4 |
=-
| 3 |
| 4 |
=
| 9 |
| 2 |
(2)原式=(-
| 9 |
| 5 |
| 25 |
| 9 |
| 3 |
| 8 |
| 1 |
| 8 |
| 1 |
| 4 |
=-5+(-
| 3 |
| 8 |
| 1 |
| 8 |
| 2 |
| 8 |
=-5+(-
| 3 |
| 8 |
| 8 |
| 3 |
=-5+1
=-4;
(3)原式=(-8)×
| 4 |
| 9 |
| 4 |
| 9 |
=-
| 32 |
| 9 |
| 4 |
| 9 |
=-4;
(4)原式=1-[
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 12 |
=1-[-6+4+1]
=1+1
=2;
(5)原式=(
| 1 |
| 3 |
| 5 |
| 21 |
| 3 |
| 14 |
| 2 |
| 7 |
| 5 |
| 6 |
=
| 1 |
| 3 |
| 5 |
| 21 |
| 3 |
| 14 |
| 2 |
| 7 |
| 5 |
| 6 |
=14-10+9-12+35
=36;
(6)原式=-9-16÷(-8)+[(-
| 1 |
| 2 |
=-9+2+1-1
=-7.
点评:本题考查的是有理数的混合运算,即先算乘方,再算乘除,最后算加减;同级运算,应按从左到右的顺序进行计算;如果有括号,要先做括号内的运算.在解答此类题目时要注意运算律的应用.
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