1/12=(1/x)+(1/y)+(1/z),xyz是三个不同的自然数,问三个数是什么?

x.y都是自然数,且x(x-u)-y(y-x)+12,求x.y得值快1且x(x-y)-y(y-x)=12 [(y*y*y-y)/(xy+1)*(xy+1)-(x+y)*(x+y)]/[(x+y)*y/(x+1)]其中x=-11,y=1/12,化简再求值. y'/y=1/x （x+y)(x+y-1)-12因式分解 y'-(1/x)y=xe^-x 解方程组 2x-y=5 2y/x-x/y=12x-y=52y/x-x/y=1 已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2) 已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2) y(x+y)+(x-y)²-（x+y）（x-y）,其中x=-1/3,y=3 （x-y)(x+y)-(x+y)^2+2y(y-x),其中x=1,y=3. 2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2 x>=1,x-y y=x-1(x (x+y)(x-y)+4(y-1) 已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 4/X+1/Y=2,XY=-1,求代数式4Y-X/X-2Y/（x+2y-12Y²/x-2y)的值 y=[x-1],