新疆卫视在线直播:一道数学题 急用

来源:百度文库 编辑:杭州交通信息网 时间:2024/05/10 13:03:02
计算,应用指数律化简
(x-y)^2*(x-y)^3
(2x+3y)^m(2x+3y)^n
[(2x-y)^2]^5
[(2x-1)(2x+3)]^3

(x-y)^2*(x-y)^3 =(x-y)^5
(2x+3y)^m(2x+3y)^n =(2x+3Y)^(m+n)
〔(2x-y)^2〕^5 =(2x-y)^10
〔(2x-1)(2x+3)〕^3 =(2x-1)^3*(2x+3)^3

(x-y)^2*(x-y)^3 =(x-y)^5
(2x+3y)^m(2x+3y)^n= (2x+3y)^(n+m)
〔(2x-y)^2〕^5 =(2x-y)^10
〔(2x-1)(2x+3)〕^3=(2x-1)^3(2x+3)^3

(x-y)^2*(x-y)^3 =(x-y)^5
(2x+3y)^m(2x+3y)^n= (2x+3y)^(n+m)
(2x-y)^2〕^5 =(2x-y)^10
(2x-1)(2x+3)〕^3=(2x-1)^3(2x+3)^3