Question

1) Tìm GTNN của bt :

A=(x-1)(2x-1)(2x²-3x-1)+2018

Answer 1

\(A=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2018\)

\(=\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2018\)

\(=\left(2x^2-3x\right)^2-1+2018\)

\(=\left(2x^2-3x\right)^2+2017\ge2017\)

\(minA=2017\Leftrightarrow2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)