三元二次方程求解(xy-y)/(xy-1)=2(x-1)(z+1)/(x+z)=3(yz+y)/(y+z+1)=4
2019-05-28
三元二次方程求解
(xy-y)/(xy-1)=2
(x-1)(z+1)/(x+z)=3
(yz+y)/(y+z+1)=4
优质解答
倒数,得(x+y-1)/（xy-y)=1/2,(x+z)/(x-1)(z+1)=1/3,(y+z+1)/(yz+y)=1/4
变形,得（（x-1)+y)/(y(x-1))=1/2,((x-1)+(z+1))/(x-1)(z+1)=1/3
(y+(z+1))/(y(z+1))=1/4
裂项,得1/y+1/(x-1)=1/2,1/(z+1)+1/(x-1)=1/3,1/(z+1)+1/y=1/4
解,得x=31/7,y=24/5,z=23.
倒数,得(x+y-1)/（xy-y)=1/2,(x+z)/(x-1)(z+1)=1/3,(y+z+1)/(yz+y)=1/4
变形,得（（x-1)+y)/(y(x-1))=1/2,((x-1)+(z+1))/(x-1)(z+1)=1/3
(y+(z+1))/(y(z+1))=1/4
裂项,得1/y+1/(x-1)=1/2,1/(z+1)+1/(x-1)=1/3,1/(z+1)+1/y=1/4
解,得x=31/7,y=24/5,z=23.