(III) với \(y\ge2\Rightarrow\left(1\right)\Leftrightarrow\frac{y-2y+4-2y+1}{y-2+y+3}=\frac{5-3y}{2y+1}< 0\Rightarrow\left[\begin{matrix}y< -\frac{1}{2}\\y>\frac{5}{3}\end{matrix}\right.\)
Kết luận(III) taco: \(\frac{5}{3}< \frac{6}{3}=2\) \(\Rightarrow y\ge2\Rightarrow x\ge4\)
