a

à lộn

\(P=x^2+2y^2+2xy-6x-8y+2027\\ =\left(x^2+y^2+9+2xy-6x-6x\right)+\left(y^2-2y+1\right)+2017\\ =\left(x+y-3\right)^2+\left(y-1\right)^2+2017\)

Do \(\left(x+y-3\right)^2\ge0\forall x;y\)

\(\left(y-1\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\forall x;y\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x+y-3\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(P_{\left(Min\right)}=2017\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

P = x²+2y² +2xy-6x-8y+2027

=x²+2xy+y²+y²-6x-6y-2y+1+9+2017

=(x²+2xy+y²)-(6x+6y)+9+(y²-2y+1)+2017

=(x+y)²-6(x+y)+9+(y-1)²+2017

=[(x+y)²-6(x+y)+9]+(y-1)² +2017

=(x+y-3)²+(y-1)²+2017

Do (x+y-3)² \(\ge0\forall x\)

(y-1)² \(\ge0\forall x\)

=>\(\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\)

=>\(\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\)=> P\(\ge2017\)

Min P=2017 khi

y-1=0

=> y=1

x+y-3=0

=>x+1-3=0

=> x=2

Vậy GTNN của P=2017 khi y=1 và x=2

\(P=x^2+2y^2+2xy-6x-8y+2024\)

\(P=x^2+y^2+y^2+2xy-6x-6y-2y+2024\)

\(P=\left(x^2+2xy+y^2\right)-\left(6x+6y\right)+9+y^2-2y+1+2014\)

\(P=\left(x+y\right)^2-2\left(x+y\right)3+3^2+\left(y^2-2y+1\right)+2014\)

\(P=\left(x+y-3\right)^2+\left(y-1\right)^2+2014\)

\(P\ge2014\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y-3=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)

Vậy.....

P = x² + 2y² + 2xy – 6x – 8y + 2028

P = (x² + y² + 2xy) – 6(x + y) + 9 + y² – 2y + 1 + 2018

P = (x + y – 3)² + (y – 1)² + 2018 \(\ge\) 2018

=> Giá trị nhỏ nhất của P = 2018 khi x = 2; y = 1

P=x²+2y²+2xy-6x-8y+2028

=x²+2xy+y²+y²-8y+x²-6x-x²+2028

=(x²+2xy+y²)+(y²-8y+16)+(x²-6x+9)-x²+2028-16-9

=(x-y)²+(y-4)²+(x-3)²-x²+2003\(\ge2003\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-4\right)^2\ge0\\\left(x-3\right)^2\ge0\\x^2\ge0\end{matrix}\right.\) nên:

Để P=2003 thì :

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\\\left(y-4\right)^2=0\\x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-3=0\\y-4=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=4\\x=0\end{matrix}\right.\)

Vậy min P=2003\(\Leftrightarrow\left(x=y\right)\in\left\{0;4;3\right\}\)

Lời giải:

$M=(x^2+y^2+2xy)+x^2+y^2-6x-6y+11$

$=(x+y)^2+x^2+y^2-6x-6y+11$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+5$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+5\geq 0+0+0+5=5$
Vậy $M_{\min}=5$. Giá trị này đạt tại $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$