Question

[Phạm Hữu Nam chuyên Đại số ♏ (Hội....]

1) tìm giá trị nhỏ nhất

 H\(=2x^2+9y^2-6xy-6x-12y+2004\) 

\(J=x^2+xy+y^2-3x-3y+1999\)

Answer 1

[Incursion_03]

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

\(\Rightarrow2H=4x^2+18y^2-12xy-12x-24y+4008\)

             \(=\left(4x^2-12xy+9y^2\right)+9y^2-12x-24y+4008\)

             \(=\left(2x-3y\right)^2-6\left(2x-3y\right)+9+9y^2-42y+49+3950\)

             \(=\left(2x-3y-3\right)^2+\left(3y-7\right)^2+3950\ge3950\)

\(\Rightarrow2H\ge3950\)

\(\Rightarrow H\ge1975\)

Dấu "=" tại \(\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)

Answer 2

[Incursion_03]

\(J=x^2+xy+y^2-3x-3y+1999\)

   \(=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}-3x-3y+1999\)

   \(=\left(x+\frac{y}{2}\right)^2-3\left(x+\frac{y}{2}\right)+\frac{9}{4}+3\left(\frac{y^2}{4}-\frac{y}{2}+\frac{1}{4}\right)+1996\)

    \(=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2+1996\ge1996\)

Dấu "=" tại \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)