Lời giải:
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=t\Rightarrow \left\{\begin{matrix} x=2t\\ y=3t\\ z=4t\end{matrix}\right.\)
Ta có: \(|x-y|=\frac{z^2}{12}\Leftrightarrow |2t-3t|=\frac{16t^2}{12}\)
\(\Leftrightarrow 3|-t|=4t^2\)
Nếu \(t\geq 0\Rightarrow 4t^2=3|-t|=3t\)
\(\Leftrightarrow t(4t-3)=0\Leftrightarrow \left[\begin{matrix} t=0\\ t=\frac{3}{4}\end{matrix}\right.\)
+) \(t=0\rightarrow x=y=z=0\rightarrow yz-x=0\)
+) \(t=\frac{3}{4}\Rightarrow x=\frac{3}{2}; y=\frac{9}{4}; z=3\) \(\rightarrow yz-x=\frac{21}{4}\)
Nếu \(t<0\Rightarrow 4t^2=3|-t|=-3t\)
\(\Leftrightarrow t(4t+3)=0\Leftrightarrow t=-\frac{3}{4}\)
\(\Rightarrow x=\frac{-3}{2}; y=\frac{-9}{4}; z=-3\rightarrow yz-x=\frac{33}{4}\)
Từ các TH trên suy ra \((yz-x)_{\max}=\frac{33}{4}\)\
\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\left|x-y\right|=\dfrac{z^2}{12}\end{matrix}\right.\) sử dụng t/c dãy tỷ bằng nhau
\(z=0\Rightarrow x=y=0=>yz-x=0\)
\(z\ne0\Rightarrow\dfrac{yz-x}{3z-2}=\dfrac{z}{4}\Rightarrow yz-x=\dfrac{z}{4}\left(3z-2\right)=\dfrac{3z^2-2z}{4}\) (1)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x-y}{-1}=\dfrac{z}{4}\Rightarrow\left|x-y\right|=\dfrac{\left|z\right|}{4}=\dfrac{z^2}{12}\)\(\Rightarrow\left[{}\begin{matrix}z=0\\z=\pm3\end{matrix}\right.\)(2)
(1) và (2) =>\(Max\left(yz-x\right)=\dfrac{3.\left(-3\right)^2-2\left(-3\right)}{4}=\dfrac{33}{4}\)