\(\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{{x^2}}}{{{{(y + 1)}^2}}} + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} = 2(1)\\
3xy = x + y + 1(2)
\end{array} \right.\\
(1) \Leftrightarrow \dfrac{{{x^2}}}{{{{(y + 1)}^2}}} - 1 + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} - 1 = 0\\
\Leftrightarrow \left( {\dfrac{x}{{y + 1}} - 1} \right)\left( {\dfrac{x}{{y + 1}} + 1} \right) + \left( {\dfrac{y}{{x + 1}} - 1} \right)\left( {\dfrac{y}{{x + 1}} + 1} \right) = 0\\
\Leftrightarrow \dfrac{{x - y - 1}}{{y + 1}}\frac{{x + y + 1}}{{y + 1}} + \dfrac{{y - x - 1}}{{x + 1}}\dfrac{{y + x + 1}}{{x + 1}} = 0\\
\Leftrightarrow \left( {x + y + 1} \right)\left[ {\dfrac{{x - y - 1}}{{{{(y + 1)}^2}}} + \dfrac{{y - x - 1}}{{{{(x + 1)}^2}}}} \right] = 0\\
\Rightarrow \left[ \begin{array}{l}
x + y + 1 = 0(*)\\
\dfrac{{x - y - 1}}{{{{(y + 1)}^2}}} + \dfrac{{y - x - 1}}{{{{(x + 1)}^2}}} = 0
\end{array} \right.(**)
\end{array}\)
Thay (*) vào (2) ta có:
\(\begin{array}{l}
(2) \Rightarrow 3xy = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
y = 0
\end{array} \right.\\
x = 1 \Rightarrow (1) \Rightarrow \dfrac{1}{{{{(y + 1)}^2}}} + \dfrac{{{y^2}}}{{{{(1 + 1)}^2}}} = 2\\
\Rightarrow 4 + {y^2}{(y + 1)^2} = 8{(y + 1)^2}
\end{array}\)
\( \Rightarrow (y + 2)({y^3} - 7x - 2) = 0\)
Suy ra có 1 nghiệm y=2
