Sửa đề: \(3\left(x-1\right)^2+2\left(y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(3\left(x-1\right)^2>=0\forall x\)
\(2\left(y-3\right)^2>=0\) với mọi y
\(\left(z+2\right)^2>=0\forall z\)
DO đó: \(3\left(x-1\right)^2+2\left(y-3\right)^2+\left(z+2\right)^2>=0\)
Dấu '=' xảy ra khi x=1;y=3; z=-2