\(x^2+y^2+4=2xy+4x+4y\)

\(\Leftrightarrow x^2-\left(2y+4\right)x+y^2-4y+4=0\)

Xét phương trình theo nghiệm x.

\(\Rightarrow\Delta'=\left(y+2\right)^2-\left(y^2-4y+4\right)=8y\)

\(\Rightarrow\orbr{\begin{cases}x=y+2-2\sqrt{2y}\\x=y+2+2\sqrt{2y}\end{cases}}\)

Vì x, y nguyên dương nên 

\(\Rightarrow\sqrt{2y}=a\)

\(\Rightarrow y=2n^2\)

\(\Rightarrow\orbr{\begin{cases}x=2n^2+2-4n\\x=2n^2+2+4n\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(n-1\right)^2\\x=2\left(n+1\right)^2\end{cases}}\)

Vậy \(\frac{y}{2};\frac{x}{2}\)là 2 số chính phương.

\(x^2+y^2+4=2xy+4x+4y\)

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

<=> \(\left(x-y-2\right)^2=8y\)

<=> \(\left(\frac{x-y-2}{4}\right)^2=\frac{y}{2}\)

=> \(\frac{y}{2}\)là số chính phương

CMTT x/2 là số chính phương

Ta có: \(4x=3y\) hay \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)

\(4y=3z\) hay \(\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{12}=\dfrac{z}{16}\left(2\right)\)

Từ (1) và (2), suy ra:

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\) \(\Rightarrow\dfrac{2x}{18}=\dfrac{y}{12}=\dfrac{z}{16}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{18}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{2x+y-z}{18+12-16}=\dfrac{-14}{14}=-1\)

Do đó:

\(\dfrac{x}{9}=-1\Rightarrow x=9.\left(-1\right)=-9\)

\(\dfrac{y}{12}=-1\Rightarrow y=12.\left(-1\right)=-12\)

\(\dfrac{z}{16}=-1\Rightarrow z=16.\left(-1\right)=-16\)

Vậy x = -9 ; y = -12 ; z = -16

\(4\left(xy+yz+xz\right)+x+y+z=9\)

Mặt khác ta có \(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\Rightarrow xy+yz+xz\le\dfrac{1}{3}\left(x+y+z\right)^2\)

\(\Rightarrow\dfrac{4}{3}\left(x+y+z\right)^2+\left(x+y+z\right)\ge9\)

\(\Leftrightarrow\left[2\left(x+y+z\right)+\dfrac{3}{4}\right]^2\ge\dfrac{441}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2\left(x+y+z\right)+\dfrac{3}{4}\ge\dfrac{21}{4}\\2\left(x+y+z\right)+\dfrac{3}{4}\le\dfrac{-21}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+z\ge\dfrac{9}{4}\\x+y+z\le-3\end{matrix}\right.\) \(\Rightarrow\left(x+y+z\right)^2\ge\dfrac{81}{16}\)

Mà \(P=x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\ge\dfrac{81}{16.3}=\dfrac{27}{16}\)

\(\Rightarrow P_{min}=\dfrac{27}{16}\) khi \(x=y=z=\dfrac{3}{4}\)

dung x^2+y^2>=2xy; x^2+1>=2x

Ta có: x²+2xy+4x+4y+3y²+3=0

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(4x+4y\right)+2y^2+3=0\)

\(\Leftrightarrow[\left(x+y\right)^2+4\left(x+y\right)+4]+2y^2=1\)

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

Do \(y^2\ge0\Rightarrow1-2y^2\le1\)

\(\Rightarrow B^2=\left(x+y+2\right)^2\le1\)

\(\Rightarrow\left\{{}\begin{matrix}B\le1\\B\ge-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}B_{max}=1\\B_{min}=-1\end{matrix}\right.\)

\(x^2+2xy+4x+4x+3y^2+3=0\\ \Leftrightarrow\left(x+y\right)^2+2.\left(x+y\right).2+4=1-2y^2\\ \Leftrightarrow\left(x+y+2\right)^2=1-2y^2\le1\\ \Rightarrow\left(x+y+2\right)^2\le1\)

\(\Rightarrow-1\le x+y+2\le1\\ \)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

\(P=\frac{1}{4x^2+2}+\frac{1}{4y^2+2}+\frac{1}{6xy}+\frac{1}{6xy}+\frac{5}{3xy}\)

\(P\ge\frac{16}{4x^2+4y^2+12xy+4}+\frac{5}{3xy}=\frac{16}{4\left(x+y\right)^2+4xy+4}+\frac{5}{3xy}\)

\(P\ge\frac{16}{4\left(x+y\right)^2+\left(x+y\right)^2+4}+\frac{5}{3.\frac{1}{4}\left(x+y\right)^2}=\frac{7}{3}\)

\(P_{min}=\frac{7}{3}\) khi \(x=y=1\)