Bài 1:
\(A=\left|3x-2\right|+\left|5-3x\right|\ge\left|3x-2+5-3x\right|=3\)
\(\Rightarrow A_{min}=3\) khi \(\frac{2}{3}\le x\le\frac{5}{3}\)
Bài 2:
Đặt \(t=\frac{2x+1}{x-3}\Rightarrow t\left(x-3\right)=2x+1\Rightarrow tx-3t=2x+1\)
\(\Rightarrow x\left(t-2\right)=3t+1\Rightarrow x=\frac{3t+1}{t-2}\) (\(t\ne2\))
Thay vào bài toán ta được:
\(f\left(t\right)=\frac{\frac{3t+1}{t-2}+2}{\frac{3t+1}{t-2}-2}=\frac{3t+1+2\left(t-2\right)}{3t+1-2\left(t-2\right)}=\frac{5t-3}{t+5}\)
Vậy \(f\left(x\right)=\frac{5x-3}{x+5}\)