Ta có :
\(P=\left(x-1\right)\left(2x+3\right)=2x^2-2x+3x-3\) \(=2x^2+x-3\)
\(=2\left(x^2+\frac{1}{2}x-\frac{3}{2}\right)\) \(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}\right)\)
\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{23}{16}\right)\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{23}{8}\ge-\frac{23}{8},\)với mọi x
Vậy \(MIN_P=\frac{-23}{8}\) khi \(x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{4}\)