题目内容
(1)-2÷(-2
)×(-4.5)
(2)-24×(
-
+
)
(3)2-4÷
×|-
|
(4)(-
+
-
)÷
(5)1÷(-3)×(-
)
(6)(-3)2-(-1
)3×
-6÷(-
)2
(7)-22×{[4
÷(-4)+(-0.4)]÷(-
)}
(8)3a+(-8a+2)-(3a-4)
(9)(x2+2xy+y2)-(x2-xy+y2)
(10)9a2-[7a2-2a-2(a2-3a)]-3.
| 1 |
| 4 |
(2)-24×(
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 12 |
(3)2-4÷
| 1 |
| 2 |
| 1 |
| 8 |
(4)(-
| 5 |
| 6 |
| 2 |
| 3 |
| 3 |
| 4 |
| 1 |
| 12 |
(5)1÷(-3)×(-
| 1 |
| 3 |
(6)(-3)2-(-1
| 1 |
| 2 |
| 2 |
| 9 |
| 2 |
| 3 |
(7)-22×{[4
| 2 |
| 3 |
| 1 |
| 3 |
(8)3a+(-8a+2)-(3a-4)
(9)(x2+2xy+y2)-(x2-xy+y2)
(10)9a2-[7a2-2a-2(a2-3a)]-3.
分析:(1)先把假分数和小数化为带分数,再按照从左到右的顺序进行计算即可;
(2)先用-24同括号中的每一项分别相乘,再把所得结果相加减即可;
(3)去绝对值符号,同时把除法变成乘法,再算乘法,最后算减法;
(4)把除法变成乘法,再根据乘法的分配律展开,最后合并即可;
(5)把除法变成乘法,再根据乘法的法则计算即可;
(6)先算乘方,再算乘除,最后算加减;
(7)先算乘方和中括号内的除法,再算括号内加法,把除法把乘法,再按乘法法则进行计算即可;
(8)先去括号,再合并同类项即可;
(9)先去括号,再合并同类项即可;
(10)先去小括号,再去中括号,最后合并同类项即可.
(2)先用-24同括号中的每一项分别相乘,再把所得结果相加减即可;
(3)去绝对值符号,同时把除法变成乘法,再算乘法,最后算减法;
(4)把除法变成乘法,再根据乘法的分配律展开,最后合并即可;
(5)把除法变成乘法,再根据乘法的法则计算即可;
(6)先算乘方,再算乘除,最后算加减;
(7)先算乘方和中括号内的除法,再算括号内加法,把除法把乘法,再按乘法法则进行计算即可;
(8)先去括号,再合并同类项即可;
(9)先去括号,再合并同类项即可;
(10)先去小括号,再去中括号,最后合并同类项即可.
解答:解:( )-2÷(-2
)×(-4.5)
=-2×(-
)×(-
)
=-4;
(2)-24×(
-
+
)
=-24×
-(-24)×
+(-24)×
=-18+20-14
=-12;
(3)2-4÷
×|-
|
=2-4×2×
=2-1
=1;
(4)(-
+
-
)÷
=(-
+
-
)×12
=-
×12+
×12-
×12
=-10+8-9
=-11;
(5)1÷(-3)×(-
)
=1×(-
)×(-
)
=
;
(6)(-3)2-(-1
)3×
-6÷(-
)2
=9-(-
)×
-6÷
=9-(-
)-6×
=9+
-
=-
;
(7)-22×{[4
÷(-4)+(-0.4)]÷(-
)}
=-4×{[
×(-
)+(-
)]×(-3)}
=-4×{[(-
)+(-
)]×(-3)}
=-4×{-
×(-3)}
=-4×
=-
;
(8)3a+(-8a+2)-(3a-4)
=3a-8a+2-3a+4
=-8a+6;
(9)(x2+2xy+y2)-(x2-xy+y2)
=x2+2xy+y2-x2+xy-y2
=3ay;
(10)9a2-[7a2-2a-2(a2-3a)]-3
=9a2-[7a2-2a-2a2+6a]-3
=9a2-[5a2+4a]-3
=9a2-5a2-4a-3
=4a2-4a-3.
| 1 |
| 4 |
=-2×(-
| 4 |
| 9 |
| 9 |
| 2 |
=-4;
(2)-24×(
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 12 |
=-24×
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 12 |
=-18+20-14
=-12;
(3)2-4÷
| 1 |
| 2 |
| 1 |
| 8 |
=2-4×2×
| 1 |
| 8 |
=2-1
=1;
(4)(-
| 5 |
| 6 |
| 2 |
| 3 |
| 3 |
| 4 |
| 1 |
| 12 |
=(-
| 5 |
| 6 |
| 2 |
| 3 |
| 3 |
| 4 |
=-
| 5 |
| 6 |
| 2 |
| 3 |
| 3 |
| 4 |
=-10+8-9
=-11;
(5)1÷(-3)×(-
| 1 |
| 3 |
=1×(-
| 1 |
| 3 |
| 1 |
| 3 |
=
| 1 |
| 9 |
(6)(-3)2-(-1
| 1 |
| 2 |
| 2 |
| 9 |
| 2 |
| 3 |
=9-(-
| 27 |
| 8 |
| 2 |
| 9 |
| 4 |
| 9 |
=9-(-
| 3 |
| 4 |
| 9 |
| 4 |
=9+
| 3 |
| 4 |
| 27 |
| 2 |
=-
| 15 |
| 4 |
(7)-22×{[4
| 2 |
| 3 |
| 1 |
| 3 |
=-4×{[
| 14 |
| 3 |
| 1 |
| 4 |
| 2 |
| 5 |
=-4×{[(-
| 7 |
| 6 |
| 2 |
| 5 |
=-4×{-
| 47 |
| 30 |
=-4×
| 47 |
| 10 |
=-
| 94 |
| 5 |
(8)3a+(-8a+2)-(3a-4)
=3a-8a+2-3a+4
=-8a+6;
(9)(x2+2xy+y2)-(x2-xy+y2)
=x2+2xy+y2-x2+xy-y2
=3ay;
(10)9a2-[7a2-2a-2(a2-3a)]-3
=9a2-[7a2-2a-2a2+6a]-3
=9a2-[5a2+4a]-3
=9a2-5a2-4a-3
=4a2-4a-3.
点评:本题考查了有理数的混合运算和整式的混合运算,主要考查学生的化简和计算能力.
练习册系列答案
相关题目
已知某山区的平均气温与该山的海拔高度的关系见下表:
(1)若海拔高度用x(米)表示,平均气温用y(℃)表示,试写出y与x之间的函数关系式;
(2)若某种植物适宜生长在18℃~20℃(包含18℃,也包含20℃)山区,请问该植物适宜种植在海拔为多少米的山区?
| 海拔高度(单位“米“) | 0 | 100 | 200 | 300 | 400 | … |
| 平均气温(单位“℃) | 22 | 21.5 | 21 | 20.5 | 20 | … |
(2)若某种植物适宜生长在18℃~20℃(包含18℃,也包含20℃)山区,请问该植物适宜种植在海拔为多少米的山区?