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maths

發問:

(２^2)*(3^3)+(2^3)*(3^2) (1/ab^1)^1 (1^1+2^2+3^3+4^4)^0 要列式 更新: 發錯..係(２^ -2)*(3^ -3)+(2^ -3)*(3^ -2) 同(1/ab^ -1)^ -1

最佳解答:

(２^2)*(3^3)+(2^3)*(3^2) = (２^2)*(3^2)*(3+2) = 4*9*5 = 180 (1/ab^1)^1 = (1/ab)^1 = 1/ab (1^1+2^2+3^3+4^4)^0 = (1+4+27+256)^0 = 1 2007-08-14 14:13:59 補充： OK, New questions(２^ -2)*(3^ -3) (2^ -3)*(3^ -2) = (1/22)*(1/33) (1/23)*(1/32)= (1/4)*(1/27) (1/8)*(1/9)= 1/108 1/72= 2/216 3/216= 5/216(1/ab^-1)^-1=(1/ab)^(-1*-1) Note: (x^a)^b = x^(ab)=(1/ab)^1=1/ab 2007-08-14 14:19:47 補充： The plus sign ( ) could not be displayed, it should be,(２^ -2)*(3^ -3) + (2^ -3)*(3^ -2) = (1/22)*(1/33) + (1/23)*(1/32)= (1/4)*(1/27) + (1/8)*(1/9)= 1/108 + 1/72= 2/216 + 3/216= 5/216

其他解答:

2^-2x3^-3+2^-3x3^-2 =1/2^2x1/3^3+1/2^3x1/3^2