Ta có \(xy\le\dfrac{\left(x+y\right)^2}{4}\).

Do đó ta có: \(x+y+xy=x+y-2xy+3xy\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Leftrightarrow\dfrac{1}{4}\left(x+y\right)^2-\left(x+y\right)\le0\)

\(\Leftrightarrow\left(x+y\right)\left[\dfrac{1}{4}\left(x+y\right)-1\right]\le0\)

\(\Leftrightarrow0\le x+y\le4\).

Do đó m = 0, n = 4.

Vậy m² + n² = 16. Chọn A.

Dạ, em cảm ơn

a: \(a^2+4b^2+9c^2=2ab+6bc+3ac\)

=>\(2a^2+8b^2+18c^2-4ab-12bc-6ac=0\)

=>\(a^2-4ab+4b^2+4b^2-12bc+9c_{}^2+a^2-6ac+9c^2=0\)

=>\(\left(a-2b\right)^2+\left(2b-3c\right)^2+\left(a-3c\right)^2=0\)

=>\(\begin{cases}a-2b=0\\ 2b-3c=0\\ 3c-a=0\end{cases}\Rightarrow a=2b=3c\)

\(A=\left(a-2b+1\right)^{2022}+\left(2b-3c-1\right)^{2023}+\left(3c-a+1\right)^{2024}\)

\(=\left(a-a+1\right)^{2022}+\left(2b-2b-1\right)^{2023}+\left(a-a+1\right)^{2024}\)

=1-1+1

=1

b: \(x^2+2xy+6x+6y+2y^2+8=0\)

=>\(x^2+2xy+y^2+6\left(x+y\right)+9+y^2-1=0\)

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

=>\(-y^2=\left(x+y+2\right)\left(x+y+4\right)\)

=>\(-y^2=\left(x+y+2024-2022\right)\left(x+y+2024-2020\right)\)

=>\(-y^2=\left(A-2022\right)\left(A-2020\right)\)

mà \(-y^2\le0\forall y\)

nên (A-2022)(A-2020)<=0

=>2020<=A<=2022

\(A_{\min}=2020\) khi x+y+2=0 và y=0

=>y=0 và x=-2-y=-2-0=-2

\(A\max=2022\) khi x+y+4=0 và y=0

=>y=0 và x=-y-4=-4

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 $y$)

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

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

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

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

\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

s y=0 v ạ 

\(=\left(x-y\right)^2+x=\left(25-5\right)^2+5=20^2+5=405\)

A   =   x 2   +   2 x y   +   y 2   –   4 x   –   4 y   +   1     =   ( x 2   +   2 x y   +   y 2 )   –   ( 4 x   +   4 y )   +   1     =   ( x   +   y ) 2   –   4 ( x   +   y )   +   1

Tại x + y = 3, ta có: A = 3 2 – 4.3 + 1 = -2

Đáp án cần chọn là: D

Ta có: \(A=\left(x^2+2xy+y^2\right)-4x-4y+1\)

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

\(=3^2-4\cdot3+1\)

\(=-2\)

a: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy\)

\(=x^2+2x+y^2-2y-2xy\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)\)

\(=\left(x-y\right)^2+2\left(x-y\right)=7^2+2\cdot7=49+14=63\)

\(B=x^3-3xy\left(x-y\right)-y^3-x^2+2xy-y^2\)

\(=x^3-3x^2y+3xy^2-y^3-\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^3-\left(x-y\right)^2\)

\(=7^3-7^2=343-49=294\)

b: \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=x^2+4xy+4y^2-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10=5^2-2\cdot5+10=25\)