Giải :(x²+2xy+y²)+y²-6x-8y+2024=(x+y)²-2(x+y)3+y²-2y+2024

=(x+y-3)²+(y²-2y+1)+2014=(x+y-3)²+(y-1)²+2014 >=2014

vì (x+y-3)²;(y-1)²>=0 với mọi x;y

nên Pmin=2014khi y=1;x=2

MinP=2024 nha!

2024 đó !đúng 100% luôn !

a

à lộn

thôi chịu

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

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

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  2018

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

A.

$a^2+4b^2+9c^2=2ab+6bc+3ac$

$\Leftrightarrow a^2+4b^2+9c^2-2ab-6bc-3ac=0$

$\Leftrightarrow 2a^2+8b^2+18c^2-4ab-12bc-6ac=0$

$\Leftrightarrow (a^2+4b^2-4ab)+(a^2+9c^2-6ac)+(4b^2+9c^2-12bc)=0$

$\Leftrightarrow (a-2b)^2+(a-3c)^2+(2b-3c)^2=0$

$\Rightarrow a-2b=a-3c=2b-3c=0$

$\Rightarrow A=(0+1)^{2022}+(0-1)^{2023}+(0+1)^{2024}=1+(-1)+1=1$

B.

$x^2+2xy+6x+6y+2y^2+8=0$

$\Leftrightarrow (x^2+2xy+y^2)+y^2+6x+6y+8=0$

$\Leftrightarrow (x+y)^2+6(x+y)+9+y^2-1=0$

$\Leftrightarrow (x+y+3)^2=1-y^2\leq 1$ (do $y^2\geq 0$ với mọi $y$)

$\Rightarrow -1\leq x+y+3\leq 1$

$\Rightarrow -4\leq x+y\leq -2$

$\Rightarrow 2020\leq x+y+2024\leq 2022$

$\Rightarrow A_{\min}=2020; A_{\max}=2022$

\(A=2x^2+y^2+2xy-6x-2y+10\)

<=>\(A=y^2+2y\left(x-1\right)+2x^2-6x+10\)

<=>\(A=y^2+2y\left(x-1\right)+\left(x^2-2x+1\right)+\left(x^2-4x+4\right)+5\)

<=>\(A=y^2+2y\left(x-1\right)+\left(x-1\right)^2+\left(x-2\right)^2+5\)

<=>\(A=\left(y+x-1\right)^2+\left(x-2\right)^2+5\ge5\)

=> A đạt giá trị nhỏ nhất là 5 khi \(\hept{\begin{cases}\left(y+x-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y+x-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}\)