đề bài này đúng ko bạn : x² -2xy + 6y²-12x+2y+45

ko đúng bn ơi 

A = x2 - 2xy +6y2 - 12x + 2y +45 

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+x^2+6x+9+1978\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x+3\right)^2+1978\)

\(=\left(x-y+1\right)^2+\left(x+3\right)^2+1978\ge1978\)

\(A_{min}=1978\) khi \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Ta có: \(C=x^2+2xy+2y^2+4y+2y+5\)

\(=x^2+2xy+y^2+y^2+6y+9-4\)

\(=\left(x+y\right)^2+\left(y+3\right)^2-4\ge-4\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x+y=0\\ y+3=0\end{cases}\Rightarrow\begin{cases}x=-y\\ y=-3\end{cases}\Rightarrow\begin{cases}x=-\left(-3\right)=3\\ y=-3\end{cases}\)

\(S=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-4y+4\right)+2021\)

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

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-3;2\right)\)

\(H=x^2+2y^2-2xy+6y+2023\\=(x^2-2xy+y^2)+(y^2+6y+9)+2014\\=(x-y)^2+(y^2+2\cdot y\cdot3+3^2)+2014\\=(x-y)^2+(y+3)^2+2014\)

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

              \(\left(y+3\right)^2\ge0\forall y\)

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

\(\Rightarrow H=\left(x-y\right)^2+\left(y+3\right)^2+2014\ge2014\forall x;y\)

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

\(\Leftrightarrow x=y=-3\)

Vậy \(Min_H=2014\) khi \(x=y=-3\)

\(H=x^2+2y^2-2xy+6y+2023\)

\(2H=2x^2+4y^2-4xy+12y+4046\)

\(2H=4y^2-4y\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)^2+2x^2+4046\)

\(2H=\left(2y-x+3\right)^2+x^2+6x+9+4028\)

\(H=\dfrac{1}{2}\left[\left(2y-x+3\right)^2+\left(x+3\right)^2\right]+2014\)

Vì \(\left(2y-x+3\right)^2+\left(x+3\right)^2\ge0\forall x,y\)

\(MinH=2014\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)

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

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

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

\(=\left(x-y-1\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

\(Vì\) \(\left(x-y-1\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0\forall x,y\)

\(MinP=\dfrac{7}{4}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Ai giúp mik với ạ 😢😭😭😭😭😢😷

\(A=\left(x^2+y^2+36-2xy-12x+12y\right)+5y^2-10y+5+109\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+109\ge109\)

\(A_{min}=109\) khi \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)