Lời giải:

$P=(x^2+y^2+2xy)+y^2-6x-8y+2028$

$=(x+y)^2-6(x+y)+(y^2-2y)+2028$
$=(x+y)^2-6(x+y)+9+(y^2-2y+1)+2018$

$=(x+y-3)^2+(y-1)^2+2018\geq 0+0+2018=2018$

Vậy $P_{\min}=2018$

Giá trị này đạt tại $x+y-3=y-1=0$

$\Leftrightarrow y=1; x=2$

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

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

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

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

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

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

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.....

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 !

\(B=\left(x^2+y^2+4+2xy-4x-4y\right)+\left(x^2+z^2+1+2xz-2x-2z\right)+\left(y^2-4y+4\right)+4\)

\(B=\left(x+y-2\right)^2+\left(x+z-1\right)^2+\left(y-2\right)^2+4\ge4\)

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

Lời giải:

$x^2+2y^2-2xy+10x-16y+20$
$=(x^2-2xy+y^2)+y^2+10x-16y+20$

$=(x-y)^2+10(x-y)+y^2-6y+20$

$=(x-y)^2+10(x-y)+25+(y^2-6y+9)-14$

$=(x-y+5)^2+(y-3)^2-14$

$\geq -14$

Vậy biểu thức có min $=-14$

Giá trị này đạt tại $x-y+5=y-3=0$

$\Leftrightarrow (x,y)=(-2,3)$

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 $y$)

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

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

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

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

\(F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=1\)

Vậy \(F_{min}=2021\)

\(\Rightarrow F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ \Rightarrow F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

 P=2x²+y²-2xy-6x+2y+2024

=>2P=4x²+2y²-4xy-12x+4y+4048

=(2x-y-3)²+y²-2y+1+4038

=(2x-y-3)²+(y-1)²+4038> hoặc = 4038

Dấu = xảy ra <=>2x-y-3=0 và y-1=0=>x=2;y=1=>2p=4038=>p=2019

Vậy Pmin=2019<=>x=2;y=1

Ta có: 

P = 2x² + y² - 2xy - 6x + 2y + 2024

P = (x² - 2xy + y²) - 2(x - y) + 1 + (x² - 4x + 4) + 2019

P = [(x - y)² - 2(x - y) + 1] + (x - 2)² + 2019

P = (x - y - 1)² + (x - 2)² + 2019 \(\ge\)2019 \(\forall\)x;y

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

Vậy MinP = 2019 <=> x = 2 và y = 1

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

\(\Rightarrow P=\left(x^2-2xy+y^2\right)-2\left(x-y\right)+1+\left(x^2-4x+4\right)+2019\)

\(\Rightarrow P=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+\left(x-2\right)^2+2019\)

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

Ta có:

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

\(\left(x-2\right)^2\ge0\forall x\inℝ\)

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

\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2+2019\ge2019\forall x;y\inℝ\)

\(\Rightarrow P\ge2019\forall x;y\inℝ\)

Dấu "=" xảy ra:

\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-1\\x=2\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=2\end{cases}}}\)

Vậy P nhỏ nhất khi P = 2019 tại x=2;y=1

Chúc bạn học tốt nhé!