计算(每小题4分,共16分)(1)(2)-22×5-(-3)×-5÷(-)(3)4x2y-9xy2+7-4x2y+10xy2-4(4)3(2x2-xy)-4(x2-xy+3)-七年级数学

题目简介

计算(每小题4分,共16分)(1)(2)-22×5-(-3)×-5÷(-)(3)4x2y-9xy2+7-4x2y+10xy2-4(4)3(2x2-xy)-4(x2-xy+3)-七年级数学

题目详情

计算(每小题4分,共16分)
(1)      
(2)-22×5-(-3)×-5÷(-
(3)4x2y-9xy2+7-4x2y+10xy2-4        
(4)3(2x2-xy)-4(x2-xy+3)
题型:解答题难度:偏易来源:不详

答案

(1)原式="0" ---------4′  
(2)55---------4′
(3)xy2+3  ---------4′  
(4)2x2+xy-12 ---------4′
(1)直接进行有理数的加减运算即可.
(2)先进行幂的运算,然后再进行有理数的乘除运算即可.
(3)直接合并同类项即可得出答案.
(4)先去括号,然后合并同类项即可.
解:(1)(-7)+(+10)+(-1)+(-2),
=-7+10-1-2,
=0.
(2)-22×5-(-3)×-5÷(-),
=-20-15÷(-),
=-20+75,
=55.
(3)4x2y-9xy2+7-4x2y+10xy2-4,
=(4x2y-4x2y)+(-9xy2+10xy2)+7-4,
=xy2+3;
(4)3(2x2-xy)-4(x2-xy+3),
=6x2-3xy-4x2+4xy-12,
=2x2+xy-12.

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