Đặt
x-2012 = a , ta sẽ có :
P= \(a^2+\left(a+4025\right)^2\)
\(=a^2+a^2+8050a+4025^2\)
\(=2a^2+8050a+4025^2\)
\(=2\left(a^2+4025a\right)+4025^2\)
= 2( \(a^2+2\cdot\dfrac{4025}{2}\cdot a+\dfrac{4025^2}{4}\))\(-\dfrac{4025^2}{4}+4025^2\)
= \(2\left(a+\dfrac{4025}{2}\right)^2+4025^2-\dfrac{4025^2}{2}\)
\(=2\left(a+\dfrac{4025}{2}\right)^2+\dfrac{4025\left(2\cdot4025-4025\right)}{2}\)
\(=2\left(a+\dfrac{4025}{2}\right)^2+\dfrac{4025^2}{2}\ge\dfrac{4025^2}{2}\)
=> MinP = \(\dfrac{4025^2}{2}\) khi \(a+\dfrac{4025}{2}=0\Rightarrow a=-\dfrac{4025}{2}\)
Mà x -2012 = \(-\dfrac{4025}{2}\Rightarrow x=2012-\dfrac{4025}{2}=-\dfrac{1}{2}\)
Vậy GTNN của P = \(\dfrac{4025^2}{2}\) khi x = \(-\dfrac{1}{2}\)
