Ta có: \(x^2+4y^2+z^2-2a+8y-6z+15\)
\(=\left(x^2-2a+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(a-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\) (Vì \(\left(a-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\forall x;y;z)\)
Vậy không có giá trị x;y;z thỏa mãn đề bài cho (đpcm)
Ta có \(x^2+4y^2+z^2-2x+8y-6z+15=0\)
<=> \(x^2-2x+1+4y^2+8y+4+z^2-6z+9+1=0\)
<=> \(\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1=0\)
<=> \(\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2=-1\)
Mà \(\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^3\ge0\forall x,y,z\) nên vô lí
Vậy....
\(x^2+4y^2+z^2-2x+8y-6z+15=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1=0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y+2\right)^2+\left(z-3\right)^2+1=0\)
Do \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)với mọi x,y,z nên
\(\left(x-1\right)^2+4\left(y-2\right)^2+\left(z-3\right)^2+1>0\)
Vậy không có x,y,z thoả mãn