Question

数学题 急！：已知x+y+z=x^2+y^2+z^2=2，求证:x/［(1-y)^2］=y/［(1-x)^2］

Answer 1

x+y+z=x^2+y^2+z^2=2
z=2-(x+y)
代入x^2+y^2+z^2=2得
x^2+y^2+[2-(x+y)]^2=2
x^2+y^2+4-4(x+y)+(x+y)^2=2
x^2+y^2+2-4(x+y)+x^2+y^2+2xy=0
x^2+y^2+1-2(x+y)+xy=0
而x/［(1-y)^2］-y/［(1-x)^2］
=[x(1-x)^2-y(1-y)^2]/[(1-x)^2(1-y)^2]
=[x(1-2x+x^2)-y(1-2y+y^2)]/[(1-x)^2(1-y)^2]
=(x-2x^2+x^3-y+2y^2-y^3)/[(1-x)^2(1-y)^2]
=[x-y-2(x^2-y^2)+x^3-y^3]/[(1-x)^2(1-y)^2]
=(x-y)[1-2(x+y)+x^2+xy+y^2]/[(1-x)^2(1-y)^2]
因为由上面得x^2+y^2+1-2(x+y)+xy=0(即上面式子的分子为0)
所以整个为0
即有x/［(1-y)^2］=y/［(1-x)^2］