题目内容
计算: (1 )
(2)
(3)
(4)
(2)
(3)
(4)
解:(1)
(2)原式=
(3)原式=12a5b6c4÷(-3a2b3c)÷[2a3b3c3]=(-4a3b3c3)÷(2a3b3c3)=-2
(4)原式=[a+(b-c)][a-(b-c)]=a2-(b-c)2=a2-(b2-2bc+c2) =a2-b2-c2+2bc
(2)原式=
(3)原式=12a5b6c4÷(-3a2b3c)÷[2a3b3c3]=(-4a3b3c3)÷(2a3b3c3)=-2
(4)原式=[a+(b-c)][a-(b-c)]=a2-(b-c)2=a2-(b2-2bc+c2) =a2-b2-c2+2bc
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