Question

tìm x,y,z biết x² + y² + z² = 4x - 2y + 6z - 14 

Answer 1

\(\Leftrightarrow x^2+y^2+z^2-4x+2y-6z+14=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+\left(z^2-6z+9\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)

Answer 2

\(\Leftrightarrow x^2+y^2+z^2-4x+2y-6z+14=0\)

\(\Leftrightarrow x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)

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

\(\Leftrightarrow x-2=0;y+1=0;z-3=0\)

\(\Leftrightarrow x=2;y=-1;z=3\)