Question

[Nguyễn Ngân Yến]

Tìm x,y,z biết : x²+3y²+z²+2xy-2yz-2x+4y+10=0

Answer 1

[Nguyễn Việt Hùng]

x^2+2xy+y^2+y^2-2yz+z^2+y^2+4y+4+6-2x=0

(x+y)^2+(y-z)^2+(y+2)^2+2*(3-x)=0

y+2=0=>y=-2

y-z=0=>z=-2 

x+y=0=>x=2

Answer 2

[ST]

<=>(x²+2xy+y²)+(y²-2yz+z²)+(y²+6y+9)-(2x+2y)+1=0

<=>[(x+y)²-2(x+y)+1]+(y-z)²+(y+3)²=0

<=>(x+y-1)²+(y-z)²+(y+3)²=0

Vì \(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}\Rightarrow\left(x+y-1\right)^2+\left(y-z\right)^2+\left(y+3\right)^2\ge0}\)

\(\Rightarrow\hept{\begin{cases}x+y-1=0\\y-z=0\\y+3=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=1\\y-z=0\\y=-3\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\z=-3\\y=-3\end{cases}}}\)

Vậy x=4,y=z=-3