若|x+y+1|与(x-y-2)2互为相反数，刚(3x-y)3的值为？（求解答过程）

Answer 1

|x+y+1|>=0(x-y-2)^2>=0面它们互为相反数，则它们只能均为0即x+y+1=0x-y-2=0两式相加得2x-1=0，x=1/2两式相减得2y+1=0，y=-1/2(3x-y)^3=(3/2+1/2)^3=2^3=8

Answer 2

若|x+y+1|与(x-y-2)2互为相反数==> x+y+1 =0, x -y - 2 = 0==> x = 1/2, y = -3/2 (3x - y)^3= [3 * (1/2)- (-3/2)]^3= [3/2 + 3/2]^3= 3^3= 27

Answer 3

|x+y+1|=-[2(x-y-2)] x+y+1=-2x+2y+4 3x-y=3 则3(3x-y)=3*3=9