题目内容
计算:(1)-22÷(-
1 |
5 |
(2)(-
1 |
2 |
1 |
3 |
1 |
4 |
(3)(8xy-x2+y2)-4(x2-y2+2xy-3);
(4)(3x2y+5xy2)-9x2y-(6x2y+2xy2-12x2y).
分析:(1)根据有理数乘除混合运算的顺序,按照从左到右的顺序依次进行运算即可;
(2)利用乘法分配律进行运算;
(3)先根据去括号法则去掉括号,再合并同类项法则合并同类项即可;
(4)先根据去括号法则去掉括号,再合并同类项法则合并同类项即可.
(2)利用乘法分配律进行运算;
(3)先根据去括号法则去掉括号,再合并同类项法则合并同类项即可;
(4)先根据去括号法则去掉括号,再合并同类项法则合并同类项即可.
解答:解:(1)-22÷(-
)2×|-5|×(-0.1)3
=-4×25×5×(-0.001)
=500×0.001
=0.5;
(2)(-
-
+
)×(-12)
=(-
)×(-12)-
×(-12)+
×(-12)
=6+4-3
=10-3
=7;
(3)(8xy-x2+y2)-4(x2-y2+2xy-3)
=8xy-x2+y2-4x2+4y2-8xy+12
=(-1-4)x2+(1+4)y2+(8-8)xy+12
=-5x2+5y2+12;
(4)(3x2y+5xy2)-9x2y-(6x2y+2xy2-12x2y)
=3x2y+5xy2-9x2y-6x2y-2xy2+12x2y
=(5-2)xy2+(3-9-6+12)x2y
=3xy2.
1 |
5 |
=-4×25×5×(-0.001)
=500×0.001
=0.5;
(2)(-
1 |
2 |
1 |
3 |
1 |
4 |
=(-
1 |
2 |
1 |
3 |
1 |
4 |
=6+4-3
=10-3
=7;
(3)(8xy-x2+y2)-4(x2-y2+2xy-3)
=8xy-x2+y2-4x2+4y2-8xy+12
=(-1-4)x2+(1+4)y2+(8-8)xy+12
=-5x2+5y2+12;
(4)(3x2y+5xy2)-9x2y-(6x2y+2xy2-12x2y)
=3x2y+5xy2-9x2y-6x2y-2xy2+12x2y
=(5-2)xy2+(3-9-6+12)x2y
=3xy2.
点评:本题考查了有理数的混合运算与整式的加减运算,熟记合并同类项法则是解题的关键,去括号时要注意符号的改变.
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