Đặt t = x - 2012 

=> P = t^2 + ( t + 4025 )^2

    P = t^2 + t^2 + 8050t + 4025^2

   P = 2t^2 + 8050t + 4025^2

       = 2 ( t^2 + 4025t ) + 4025^2

         = 2 ( t^2 + 2.t.4025/2 + 4025^2/4 ) -  4025^2/2 + 4025^2 

         = 2 ( t + 4025/2 )^2 + 4025^2 - 4025^2/2 

Vậy GTNN là 4025^2 - 4025^2/2 khi t + 4025/2 = 0 => t = -4025/2 

=> x - 2012 = -4025/2 => x = ... 

ĐKXĐ: \(x-2013\ge0\Leftrightarrow x\ge2013\)

Ta có:

\(A=\sqrt{x-2013-2\sqrt{x-2013}+1}+\sqrt{x-2013-90\sqrt{x-2013}+2025}\)

\(=\sqrt{\left(\sqrt{x-2013}-1\right)^2}+\sqrt{\left(\sqrt{x-2013}-45\right)^2}\)

\(=\left|\sqrt{x-2013}-1\right|+\left|\sqrt{x-2013}-45\right|\)

\(=\left|\sqrt{x-2013}-1\right|+\left|45-\sqrt{x-2013}\right|\)

\(\ge\left|\sqrt{x-2013}-1+45-\sqrt{x-2013}\right|\)

\(=\left|-1+45\right|=\left|44\right|=44\)

Vậy GTNN của A là 44, đạt được khi và chỉ khi \(\left(\sqrt{x-2013}-1\right)\left(45-\sqrt{x-2013}\right)\ge0\)

\(\Leftrightarrow1\le\sqrt{x-2013}\le45\)

\(\Leftrightarrow1\le x-2013\le2025\)

\(\Leftrightarrow2014\le x\le4038\left(tm\right)\)

Áp dụng BĐT AM-GM ta có:

\(A=\frac{2011x+2012\sqrt{1-x^2}+2013}{\sqrt{1-x^2}}\)\(=\frac{2011x+2013}{\sqrt{1-x^2}}+2012\)

\(=\frac{2012\left(x+1\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2012\)\(\ge\frac{2\sqrt{2012\left(x+1\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2012\)

\(\ge\frac{2\sqrt{2012\left(1-x^2\right)}}{\sqrt{1-x^2}}+2012=2\sqrt{2012}+2012\)

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 = \(-\dfrac{1}{2}\)

|x + 2013|  lớn hơn hoặc bằng 0,|y - 2012| lớn hơn hoặc bằng 0

=>|x + 2013| + |y - 2012| lớn hơn hoặc bằng 0

khi |x + 2013| + |y - 2012| lớn hơn hoặc bằng 0 thì x=-2013,y=2012

vậy x=-2013,y=2012 

tick nhé

Đặt P = x - 2012 

=> P = t^2 + ( t + 4025 )^2

    P = t^2 + t^2 + 8050t + 4025^2

   P = 2t^2 + 8050t + 4025^2

       = 2 ( t^2 + 4025t ) + 4025^2

         = 2 ( t^2 + 2.t.4025/2 + 4025^2/4 ) -  4025^2/2 + 4025^2 

         = 2 ( t + 4025/2 )^2 + 4025^2 - 4025^2/2 

Vậy GTNN là 4025^2 - 4025^2/2 khi t + 4025/2 = 0 => t = -4025/2 

=> x - 2012 = -4025/2 => x = ...