A= x²+2y²-2xy-2x-2y+1015

A = x² - 2xy - 2x + y² + 2y + 1 + y² - 4y + 4 + 1010 

A = [x² - 2x(y + 1) + (y+1)² ]  + (y-2)² + 1010

A = ( x - y - 1)² + (y-2)² + 1010 \(\ge1010\forall x,y\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)

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

Vậy MinA = 1010 <=> \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

\(Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5=\left[\left(x^2+2xy+y^2\right)-2z\left(x+y\right)+z^2\right]+\left(y^2-2y+1\right)+\left(z^2+4z+4\right)=\left(x+y-z\right)^2+\left(y-1\right)^2+\left(z+2\right)^2\ge0\)

\(minQ=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-3\\y=1\\z=-2\end{matrix}\right.\)

`Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5`

`Q=(x^2+y^2-z^2+2xy-2yz-2xz)+(y^2-2y+1)+(z^2+4z+4)`

`Q=(x+y-z)^2+(y-1)^2+(z+2)^2`

Ta thấy :

`(x+y-z)^2>=0`

`(y-1)^2>=0`

`(z+2)^2>=0`

`=>(x+y-z)^2+(y-1)^2+(z+2)^2>=0`

Dấu = xảy ra 

`<=>` $\begin{cases}x+y-z=0\\y-1=0\\z+2=0\end{cases}$

`<=>` $\begin{cases}x=-3\\y=1\\z=-2\end{cases}$

Bạn nên sửa lại đề là tìm GTNN

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2+4y+4+15\\ A=\left(x-y+1\right)^2+\left(y+2\right)^2+15\ge15\\ A_{min}=15\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Vậy GTNN của A là 15

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   =   x 2   +   2 y 2   –   2 x y   +   2 x   –   10 y     ⇔   A   =   x 2   +   y 2   +   1   –   2 x y   +   2 x   –   2 y   +   y 2   –   8 y   +   16   –   17     ⇔   A   =   ( x 2   +   y 2   +   12   –   2 . x . y   +   2 . x . 1   –   2 . y . 1 )   +   ( y 2   –   2 . 4 . y   +   4 2 )   –   17     ⇔   A   =   ( x   –   y   +   1 ) 2   +   ( y   –   4 ) 2   –   17

Vì  với mọi x; y nên A ≥ -17 với mọi x; y

=> A = -17 

⇔ x − y + 1 = 0 y − 4 = 0 ⇔ x = y − 1 y = 4 ⇔ x = 3 y = 4

Vậy A đạt giá trị nhỏ nhất là A = -17 tại   x = 3 y = 4

Đáp án cần chọn là: B

A   =   x 2   +   2 y 2   –   2 x y   +   2 x   –   10 y     ⇔   A   =   x 2   +   y 2   +   1   –   2 x y   +   2 x   –   2 y   +   y 2   –   8 y   +   16   –   17     ⇔   A   =   ( x 2   +   y 2   +   1 2   –   2 . x . y   +   2 . x . 1   –   2 . y . 1 )   +   ( y 2   –   2 . 4 . y   +   4 2 )   –   17     ⇔   A   =   ( x   –   y   +   1 ) 2   +   ( y   –   4 ) 2   –   17

Vì x - y + 1 2 ≥ 0 y - 4 2 ≥ 0  với mọi x, y nên A ≥ -17 với mọi x, y

=> A = -17 ó x - y + 1 = 0 y - 4 = 0 ó x = y - 1 y = 4 ó x = 3 y = 4  

Vậy A đạt giá trị nhỏ nhất là A = -17 tại   x = 3 y = 4

Đáp án cần chọn là: C

\(A=\left[\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1\right]+\left(y^2-8y+16\right)-17\\ A=\left(x-y+1\right)^2+\left(y-4\right)^2-17\ge-17\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1=3\\y=4\end{matrix}\right.\)

\(A=x^2-2xy+2y^2-4y+5\\=(x^2-2xy+y^2)+(y^2-4y+4)+1\\=(x-y)^2+(y-2)^2+1\)

Ta thấy: \(\left(x-y\right)^2\ge0\forall x;y\)

              \(\left(y-2\right)^2\ge0\forall y\)

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

\(\Rightarrow A=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\forall x;y\)

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

\(\Leftrightarrow x=y=2\)

Vậy \(Min_A=1\) khi \(x=y=2\).

$Toru$

Thêm đấu ngoặc vô đi 

với x;y>=0 ta có:

\(A^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2=2x+1+2y+1+2\sqrt{\left(2x+1\right)\left(2y+1\right)}\)

\(=2\left(x+y\right)+2+\sqrt{4xy+2x+2y+1}=2\left(x+y\right)+2+\sqrt{4xy+2\left(x+y\right)+1}\)

\(2=2\left(x^2+y^2\right)=\left(1+1\right)\left(x^2+y^2\right)>=\left(x+y\right)^2\Rightarrow x+y< =\sqrt{2}\)(bđt bunhiacopxki)

\(2xy< =x^2+y^2=1\Rightarrow2\cdot2xy=4xy< =2\cdot1=2\)

\(\Rightarrow A^2=2\left(x+y\right)+2+2\sqrt{4xy+2\left(x+y\right)+1}< =2\sqrt{2}+2+2\sqrt{2+2\sqrt{2}+1}\)

\(=2\sqrt{2}+2+2\sqrt{\left(\sqrt{2}+1\right)^2}=2\sqrt{2}+2+2\left(\sqrt{2}+1\right)4\sqrt{2}+4\)

\(\Rightarrow A< =\sqrt{4\sqrt{2}+4}\)

dấu = xảy ra khi x=y=\(\sqrt{\frac{1}{2}}\)

vậy max A là \(\sqrt{4\sqrt{2}+4}\)khi \(x=y=\sqrt{\frac{1}{2}}\)