Giải hệ phương trình : $left { {{frac{x^2}{(y+1)^2}+frac{y^2}{(x+1)^2}=frac{2}{1}\3xy=x+y+1 } atop {}} ight.$

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1. \(\begin{array}{l}
   \left\{ \begin{array}{l}
   \dfrac{{{x^2}}}{{{{(y + 1)}^2}}} + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} = 2(1)\\
   3xy = x + y + 1(2)
   \end{array} \right.\\
   (1) \Leftrightarrow \dfrac{{{x^2}}}{{{{(y + 1)}^2}}} – 1 + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} – 1 = 0\\
   \Leftrightarrow \left( {\dfrac{x}{{y + 1}} – 1} \right)\left( {\dfrac{x}{{y + 1}} + 1} \right) + \left( {\dfrac{y}{{x + 1}} – 1} \right)\left( {\dfrac{y}{{x + 1}} + 1} \right) = 0\\
   \Leftrightarrow \dfrac{{x – y – 1}}{{y + 1}}\frac{{x + y + 1}}{{y + 1}} + \dfrac{{y – x – 1}}{{x + 1}}\dfrac{{y + x + 1}}{{x + 1}} = 0\\
   \Leftrightarrow \left( {x + y + 1} \right)\left[ {\dfrac{{x – y – 1}}{{{{(y + 1)}^2}}} + \dfrac{{y – x – 1}}{{{{(x + 1)}^2}}}} \right] = 0\\
   \Rightarrow \left[ \begin{array}{l}
   x + y + 1 = 0(*)\\
   \dfrac{{x – y – 1}}{{{{(y + 1)}^2}}} + \dfrac{{y – x – 1}}{{{{(x + 1)}^2}}} = 0
   \end{array} \right.(**)
   \end{array}\)
   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

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