Question

tìm x,y,z sao cho 

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

Answer 1

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)

\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)

\(\Rightarrow x=1;y=-2;z=-1\)

Answer 2

<=>(x²-2x+1)+(3y²+12y+12)+(2z²+4z+2)=0

<=>(x-1)²+3(y+2)²+2(z+1)²=0

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

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)