Question

tìm giá trị nhỏ nhất của P= (x+1)^2+(x-2)^2+2019

Answer 1

\(P=\left(x+1\right)^2+\left(x-2\right)^2+2019\)

\(=x^2+2x+1+x^2-4x+4+2019\)

\(=2x^2-2x+2024\)

\(=2\left(x^2-x+1012\right)\)

\(=2\left[\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{4047}{4}\right]\)

\(=2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{4047}{4}\right]\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{4047}{4}\right]\ge2\times\dfrac{4047}{4}=\dfrac{4047}{2}\)

Vậy GTNN của P là \(\dfrac{4047}{2}\) . Dấu \("="\) xảy ra khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Wish you study well !!