Question

已知 2（【根号x】+【根号y－1】+【根号z－2】）=x+y+z,求x,y,z的值

Answer 1

x+y+z=(√x)^2+(√y-1)^2+1+(√(z-2))^2+2

∴

2[√x +√(y-1)+√(z-2)]=

=(√x)^2+(√y-1)^2+1+(√(z-2))^2+2

=(√x)^2+1+(√y-1)^2+1+(√(z-2))^2+1

∴

(√x)^2+1+(√y-1)^2+1+(√(z-2))^2+1-2√x -2√(y-1)-2√(z-2)=0

（√x -1）^2+(√y-1 -1)^2+(√z-2 -1)^2=0

√x =1 x=1

√y-1 =1 y=2

√z-2 =1 z=3