Question

 Bài 1:Tìm GTNN 

\(x^2+y^2-3x+2y+3\)\(2x^2+5y^2+4xy+8x-4y+15\)

Bài 2: Tìm GTLN

\(12x-4x^2+3\) \(12x-8y-4x^2-y^2-2\)

Answer 1

Khó phết chứ chả đùa

Answer 2

Bài 1:

1.Đặt \(A=x^2+y^2-3x+2y+3\)

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

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

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

Vì \(\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0;\forall x\\\left(y+1\right)^2\ge0;\forall y\end{cases}}\)

\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2\ge0;\forall x,y\)

\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{1}{4}\ge0-\frac{1}{4};\forall x,y\)

Hay \(A\ge\frac{-1}{4};\forall x,y\)

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

                       \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)

VẬY MIN A=\(\frac{-1}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)

Answer 3

Bài 2:

1.Đặt \(C=12x-4x^2+3\)

\(=-4x^2+12x+3\)

\(-4x^2+12x-9+12\)

\(=-\left(4x^2-12x+9\right)+12\)

\(=-\left(2x-3\right)^2+12\)

Vì \(-\left(2x-3\right)^2\le0;\forall x\)

\(\Rightarrow-\left(2x-3\right)^2+12\le0+12;\forall x\)

Hay \(C\le12;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x-3\right)^2=0\)

                        \(\Leftrightarrow x=\frac{3}{2}\)

Vậy MAX C=12 \(\Leftrightarrow x=\frac{3}{2}\)

Answer 4

2. Đặt \(D=12x-8y-4x^2-y^2-2\)

\(=-4x^2-y^2-8y+12x-2\)

\(=-\left(4x^2+y^2+8y-12x+2\right)\)

\(=-\left[\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)-23\right]\)

\(=-\left(2x-3\right)^2-\left(y+4\right)^2+23\)

Vì \(\hept{\begin{cases}-\left(2x-3\right)^2\le0;\forall x\\-\left(y+4\right)^2\le0;\forall y\end{cases}}\)

\(\Rightarrow-\left(2x-3\right)^2-\left(y+4\right)^2\le0;\forall x,y\)

\(\Rightarrow-\left(2x-3\right)^2-\left(y+4\right)^2+23\le0+23;\forall x,y\)

Hay \(D\le23;\forall x,y\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}\left(2x-3\right)^2=0\\\left(y+4\right)^2=0\end{cases}}\)

                     \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)

Vậy MAX D= 23 \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)

Answer 5

Bài 1:

2. Đặt \(B=2x^2+5y^2+4xy+8x-4y+15\)

\(=\left(x^2+4xy+4y^2\right)+\left(y^2-4y+4\right)+\left(x^2+8x+16\right)-5\)

\(=\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2-5\)

Vì \(\hept{\begin{cases}\left(x+2y\right)^2\ge0;\forall x,y\\\left(y-2\right)^2\ge0;\forall x,y\\\left(x+4\right)^2\ge0;\forall x,y\end{cases}}\)

\(\Rightarrow\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2-5\ge0-5;\forall x,y\)

Hay \(B\ge-5;\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(y-2\right)^2=0\\\left(x+4\right)^2=0\end{cases}}\)

                        \(\Leftrightarrow\hept{\begin{cases}y=2\\x=-4\end{cases}}\)

Vậy MIN \(B=-5\)\(\Leftrightarrow\hept{\begin{cases}y=2\\x=-4\end{cases}}\)