\(P=\left(x-1\right)\left\{2\left(x-1\right)+4\right\}=2\left(x-1\right)^2+4\left(x-1\right)\)
\(P=2y^2+4y=2\left(y+1\right)^2-2\ge-2\)
khi y=-1=> x=0
Ta có:P=(x-1)(2x+3)
P=2x2+x-3
P=2(x2+2.\(\frac{1}{4}\)+\(\frac{1}{8}\))-\(\frac{13}{4}\)
\(P=2.\left(x+\frac{1}{4}\right)^2-\frac{13}{4}\)
Vì \(2.\left(x+\frac{1}{4}\right)^2\le0\)
Suy ra:\(2.\left(x+\frac{1}{4}\right)^2-\frac{13}{4}\le-\frac{13}{4}\)
Dấu = xảy ra khi \(x+\frac{1}{4}=0\Rightarrow x=-\frac{1}{4}\)
Vậy Max P=\(-\frac{13}{4}\)khi \(x=-\frac{1}{4}\)
