\(E=4x^2+4xy+y^2+x^2-2x+1+y^2+4y+4+2005\)

\(=\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+2005\ge2005\)

\(E_{min}=2005\) khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Đặt \(P=x^2+4y^2-4xy+2x-4y+9\)

\(P=\left(x-2y\right)^2+2\left(x-2y\right)+1+8\)

\(P=\left(x-2y+1\right)^2+8\ge8\)

\(P_{min}=8\) khi \(x-2y+1=0\)

B= \(4x^2+4xy+y^2+x^2-2x+1+y^2+4y+4+15\)

= \(\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+15\ge15\)

=> GTNN của B là 15

Ta có: A = 5x² + 2y² + 4xy - 2x + 4y + 2005

             = (4x²+ 4xy+y² ) + ( x² - 2x + 1) + (y² + 4y + 2) + 2002

             = (2x+y)² + (x-1)² + (y+2)² +2002

Ta có: (2x+y)²>=0 V x,y. Dấu "=" XR khi 2x+y=0 <=> 2x=-y

          (x-1)² >=0 Vx. Dấu "=" XR khi x=1

          ((y+2)² >=0 V y. Dấu "=" XR khi y=-2

Vậy A>=2002 V x,y. Dấu "=" XR khi 2x=-y; x=1; y=2 <=> (x,y)=(1;2)

Do đó Min A=2002 tại (x,y)=(1,2)

Kẻ Vô Danh: Em kết luận giá trị y sai nhé.

GTNN của A  là 2002 khi  x = 1, y = - 2.

\(A=3x^2+5x-2\)

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

\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2-\frac{49}{36}\right)\)

\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2\right)-\frac{49}{12}\)

\(A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\)

         Vì \(3\left(x+\frac{5}{6}\right)^2\ge0\)

                  Do đó \(3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)

Dấu = xảy ra khi \(x+\frac{5}{6}=0\Rightarrow x=-\frac{5}{6}\)

      Vậy Min A=\(-\frac{49}{12}\) khi x=\(-\frac{5}{6}\)

mk làm ý a thôi, mấy ý sau dựa vào mà làm.

      A = \(3x^2+5x-2\)

 => \(\frac{A}{3}=x^2+\frac{5}{3}x-\frac{2}{3}\)(chia cả 2 vế cho 3)

\(\Leftrightarrow\frac{A}{3}=x^2+2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2-\frac{49}{36}\)

\(\Leftrightarrow\frac{A}{3}=\left(x+\frac{5}{6}\right)^2-\frac{49}{36}\)

\(\Rightarrow A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)

Đẳng thức xảy ra <=> x = - 5/6.

Vậy Min A = - 49/12 khi và chỉ khi x = - 5/6.

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)

\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)