k =(2x-y)² + (x+1)² +( y +2)² +2y² -5+3

gtnn k = -2

Ta có K = (x²  + 4y² + 1 - 4xy - 2x + 4y) + (4x² + 4x + 1) + 1 = (2y - x + 1)² + (2x + 1)² + 1 >= 1

Vậy GTNN là -1 đạt được tại x = -0,5; y = - 0,25

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

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

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

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

\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)

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

\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)

mk sửa lại đoạn sau:

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x-1=0\\2z-x-1=0\\2y+x-z-1=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\2z-2=0\\2y-z=0\left(x-1=0\right)\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\2y=1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\y=\dfrac{1}{2}\end{matrix}\right.\)

Ta có:

3x²+4y²+5z²+4xy-4yz-6xz-2x-4y-2z+3=0

<=> 4y²+4yx-4yz-4y+x²-2xz+z²-2x+2z+1+4z²-4xz-4z+x²+2x+1+x²-2x+1=0

<=>4y²+4y(x-z-1)+(x-z)²-2(x-z)+1+4z²-4z(x+1)+(x+1)²+(x-1)²=0

<=>4y²+4y(x-z-1)+(x-z-1)²+(2z-x-1)²+(x-1)²=0

<=>(2y+x-z-1)²+(2z-x-1)²+(x-1)²=0

<=>\(\left\{{}\begin{matrix}x-1=0\\2z-x-1=0\\2y+x-z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2z=0\left(x-1=0\right)\\2y-z=0\left(x-1=0\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\z=0\\y=0\end{matrix}\right.\)

Vay....