\(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge0\)
=>đpcm
Đúng 0
Bình luận (9)
x2+4y2+z2-2x-6z+8y+15
=x2+4y2+z2-2x-6z+8y+1+1+4+9
=(x2-2x+1)+(4y2+8y+4)+(z2-6z+9)+1
=(x-1)2+4(y+1)2+(z-3)2+1
Ta thấy:\(\begin{cases}\left(x-1\right)^2\\4\left(y+1\right)^2\\\left(z-3\right)^2\end{cases}\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge0+1=1>0\)
Đpcm
Đúng 0
Bình luận (0)
