Lời giải:
PT \(\Leftrightarrow |(x-3)(x+1)|=|2x-5|+1\)
Ta xét các TH sau:
\(x\geq 3\Rightarrow \left\{\begin{matrix} |(x-3)(x+1)|=(x-3)(x+1)\\ |2x-5|=2x-5\end{matrix}\right.\)
PT trở thành: \((x-3)(x+1)=2x-5+1\)
\(\Leftrightarrow x^2-2x-3=2x-5+1\)
\(\Leftrightarrow x^2-4x+1=0\)
\(\Rightarrow x=2\pm \sqrt{3}\). Vì \(x\geq 3\Rightarrow x=2+\sqrt{3}\)
TH2: \(\frac{5}{2}\leq x< 3\Rightarrow \left\{\begin{matrix} |(x-3)(x+1)|=-(x-3)(x+1)\\ |2x-5|=2x-5\end{matrix}\right.\)
PT trở thành:
\(-(x-3)(x+1)=2x-5+1\)
\(\Leftrightarrow -x^2+2x+3=2x-5+1\)
\(\Leftrightarrow -x^2+7=0\Rightarrow x= \sqrt{7}\) (vì \(\frac{5}{2}\leq x< 3\) )
TH3: \(-1\leq x< \frac{5}{2}\Rightarrow \left\{\begin{matrix} |(x-3)(x+1)|=-(x-3)(x+1)\\ |2x-5|=5-2x\end{matrix}\right.\)
PT trở thành:
\(-(x-3)(x+1)=5-2x+1\)
\(\Leftrightarrow -x^2+2x+3=6-2x\)
\(\Leftrightarrow -x^2+4x-3=0\Rightarrow x=3; x=1\). Vì \(-1\leq x< \frac{5}{2}\Rightarrow x=1\)
TH4: \(x< -1\Rightarrow \left\{\begin{matrix} |(x-3)(x+1)|=(x-3)(x+1)\\ |2x-5|=5-2x\end{matrix}\right.\)
PT trở thành:
\((x-3)(x+1)=5-2x+1\)
\(\Leftrightarrow x^2-2x-3=6-2x\)
\(\Leftrightarrow x^2-9=0\Rightarrow x=\pm 3\). Vì $x< -1$ nên $x=-3$
Vậy.............
