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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2x^2+9y^2-6xy-6x-12y+2004\)
\(A=\left(3y\right)^2-2\cdot3y\cdot2+2^2+2x^2-6x+2000\)
\(A=\left(3y-2\right)^2+2\left(x^2-2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2\right)+1997,75\)
\(A=\left(3y-2\right)^2+2\left(x-\frac{3}{2}\right)^2+1997,75\)
\(A\ge1997,75\)
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,.........
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ⇔ \(x=1\)