\(pt:2x-3=\frac{x^2-8x+15}{x-3}\) \(ĐKXĐ:x\ne3\)
\(\Rightarrow\left(2x-3\right)\left(x-3\right)=x^2-8x+15\)
\(\Leftrightarrow2x^2-6x-3x+9-x^2+8x-15=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loai\right)\\x=3\left(loai\right)\end{matrix}\right.\)
\(\varphi TH_2:2x-3< 0\Leftrightarrow x< \frac{3}{2}\)\(pt:3-2x=\frac{x^2-8x+15}{x-3}\)
\(\Rightarrow\left(3-2x\right)\left(x-3\right)=x^2-8x+15\)
\(\Leftrightarrow3x-9-2x^2+6x-x^2+8x-15=0\)
\(\Leftrightarrow-3x^2+17x-24=0\)
\(\Leftrightarrow-3x^2+8x+9x-24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\left(loai\right)\\x=3\left(loai\right)\end{matrix}\right.\)
Vậy \(S=\varnothing\)
\(pt:2x-3=\frac{x^2-8x+15}{x-3}\)
\(\Rightarrow\left(2x-3\right)\left(x-3\right)=x^2-8x+15\)
\(\Leftrightarrow2x^2-6x-3x+9-x^2+8x-15=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loai\right)\\x=3\left(nhan\right)\end{matrix}\right.\)
\(\varphi TH_2:2x-3< 0\Leftrightarrow x< \frac{3}{2}\)\(pt:3-2x=\frac{x^2-8x+15}{x-3}\)
\(\Rightarrow\left(3-2x\right)\left(x-3\right)=x^2-8x+15\)
\(\Leftrightarrow3x-9-2x^2+6x-x^2+8x-15=0\)
\(\Leftrightarrow-3x^2+17x-24=0\)
\(\Leftrightarrow-3x^2+8x+9x-24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\left(loai\right)\\x=3\left(loai\right)\end{matrix}\right.\)
Vậy \(S=\varnothing\)
