a) \(A=2x^2+9y^2-6xy-6x-12y+2014\)

\(=\left(2x^2-6xy-6x\right)+\left(9y^2-12y\right)+2014\)

\(=2\left[x^2-2.x.\frac{3\left(y+1\right)}{2}+\frac{9\left(y+1\right)^2}{4}\right]+\left[9y^2-12y-\frac{9}{2}.\left(y+1\right)^2\right]+2014\)

\(=2\left[x-\frac{3\left(y+1\right)}{2}\right]^2+\frac{1}{2}\left(3y-7\right)^2+1985\ge1985\)

Dấu "=" xảy ra khi và chỉ khi y = \(\frac{7}{3}\Rightarrow x=5\)

Vậy Min A = 1985 tại \(\left(x;y\right)=\left(5;\frac{7}{3}\right)\)

b) \(B=-x^2+2xy-4y^2+2x+10y-8\)

\(=-\left(x^2-2xy-2x\right)-\left(4y^2-10y\right)-8\)

\(=-\left[x^2-2x\left(y+1\right)+\left(y+1\right)^2\right]-\left[4y^2-10y-\left(y+1\right)^2\right]-8\)

\(=-\left(x-y-1\right)^2-\left(y-2\right)^2+5\le5\)

Dấu đẳng thức xảy ra khi và chỉ khi y = 2 => x = 3

Vậy B đạt giá trị lớn nhất bằng 5 tại (x;y) = (3;2)

để mai nhé @

a. Min A= 2014 khi x= 0, y= 0

\(A=2x^2+9y^2-6xy-6x-12y+2004\)

\(A=\left(3y\right)^2-2\cdot3y\cdot2+2^2+2x^2-6x+2000\)

\(A=\left(3y-2\right)^2+2\left(x^2-2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2\right)+1997,75\)

\(A=\left(3y-2\right)^2+2\left(x-\frac{3}{2}\right)^2+1997,75\)

\(A\ge1997,75\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3y-2=0\\x-\frac{3}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}}\)

Vậy,.........

Sửa cho Bonking ( bắt đầu dòng 3 )

\(A=\left(3y-2\right)^2+2\left(x^2-2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right)+2000\)

\(A=\left(3y-2\right)^2+2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]+2000\)

\(A=\left(3y-2\right)^2+2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}+2000\)

\(A=\left(3y-2\right)^2+2\left(x-\frac{3}{2}\right)^2+1995,5\)

\(A\ge1995,5\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3y-2=0\\x-\frac{3}{2}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)

Vậy,.........

\(A=2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(x^2-6xy-12y+9y^2+4+4x\right)+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

Dấu "=" xảy ra khi\(\hept{\begin{cases}\left(x-3y+2\right)^2=0\\\left(x-5\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}}\)

\(B=-x^2+2xy-4y^2+2x+10y-8\)

\(-B=\left(x^2+y^2+1-2xy+2y-2x\right)+3\left(y^2-4y+4\right)-5\)

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

\(B=5-[\left(x-y-1\right)^2+\left(y-2\right)^2]\le5\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

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

\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)