Ta có pt <=> \(y^2-2=3x-2xy\Leftrightarrow y^2-2=x\left(3-2y\right)\)

<=> x=\(\frac{y^2-2}{3-2y}\)

để x là số nguyên <=> \(\frac{y^2-2}{3-2y}\in Z\Leftrightarrow y^2-2⋮3-2y\)

=> \(4y^2-8⋮2y-3\Leftrightarrow4y^2-6y+6y-9+1⋮2y-3\)

<=> \(2y-3\inư\left(1\right)=\left\{\pm1\right\}\)

xét rồi thay vào nhá !!

^^

y = 1

x = -1

\(\left(x-1\right)^2\ge-4x\Leftrightarrow\left(x-1\right)^2+4x\ge0\)

\(\Leftrightarrow x^2-2x+1+4x\ge0\)

\(\Leftrightarrow x^2+2x+1\ge0\)

\(\Leftrightarrow\left(x+1\right)^2\ge0\)(luôn đúng)

\(\text{Ta có : }2x^2+\frac{1}{x^2}+\frac{y^2}{4}=4\)

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

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

\(\text{ Lại có : }\left(x+\frac{1}{x}\right)^2+\left(x-\frac{y}{2}\right)^2\ge0\)

\(\Rightarrow2-xy\ge0\)

\(\Rightarrow xy\le2\)

Mà xy có giá trị lớn nhất 

\(\Rightarrow xy\in\left\{\left(1;2\right)\left(2;1\right)\left(-1;-2\right)\left(-2;-1\right)\right\}\)

áp dụng bổ đề     \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bạn dùng cô-si,xét tích \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\))

\(\Leftrightarrow\frac{1}{x^2+2xy}+\frac{1}{y^2+2yz}+\frac{1}{z^2+2xz}\ge\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1^2}\)