Question

Cho x;y;z là các số hữu tỉ khác ) sao cho \(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\).

Tính M = \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8xyz}\).

Answer 1

TH1 : \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)

Th2 : \(x+y+z\ne0\)

\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)

\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)

\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)

\(\Leftrightarrow x=y=z\)

\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)

Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)