计算:(1)−22÷(−15)2×|−5|×(−0.1)3;(2)(−12−13+14)×(−12)(3)(8xy-x2

计算:
(1)22÷(−
1
5
)2×|−5|×(−0.1)3

(2)(−
1
2
1
3
+
1
4
)×(−12)

(3)(8xy-x2+y2)-4(x2-y2+2xy-3);
(4)(3x2y+5xy2)-9x2y-(6x2y+2xy2-12x2y).
newuser3008 1年前 已收到1个回答 举报

6nn8602 幼苗

共回答了12个问题采纳率:83.3% 举报

解题思路:(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.

(1)−22÷(−
1
5)2×|−5|×(−0.1)3
=-4×25×5×(-0.001)
=500×0.001
=0.5;

(2)(−
1
2−
1
3+
1
4)×(−12)
=(-[1/2])×(-12)-[1/3]×(-12)+[1/4]×(-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

点评:
本题考点: 整式的加减;有理数的混合运算.

考点点评: 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.

1年前

8
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