\(P=\left(x-1\right)\left(2x+3\right)=2x^2+x-3\)
\(=2\left(x^2+\frac{x}{2}-\frac{3}{2}\right)\)
\(=2\left(x^2+\frac{x}{2}+\frac{1}{16}\right)-\frac{25}{8}\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{25}{8}\ge-\frac{25}{8}\)
Dấu "=" khi \(x=-\frac{1}{4}\)
Vậy \(Min_P=-\frac{25}{8}\Leftrightarrow x=-\frac{1}{4}\)