Question

giai phuong trinh x^2+y^2+z^2=y(x+z)

Answer 1

\(PT\Leftrightarrow x^2+y^2+z^2=xy+yz\)

\(\Leftrightarrow4x^2+4y^2+4z^2=4xy+4yz\)

\(\Leftrightarrow4x^2+4y^2+4z^2-4xy-4yz=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4z^2-4yz+y^2\right)+2y^2=0\)

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

Vì \(\left(2x-y\right)^2+\left(2z-y\right)^2+2y^2\ge0\forall x;y;z\)

Dấu "=" xảy ra khi \(x=y=z=0\)