Решите систему уравнений: ((x+2)/y)-(3y/(x+2))=2; xy=16.

\[(*)\text{\ \ }\frac{(x + 2)^{2} - 3y^{2}}{y(x + 2)} = 2\]

\[x^{2} + 4x + 4 - 3y^{2} =\]

\[= 2xy + 4y\ \left( xy = 16;\ \ \ \ y = \frac{16}{x} \right)\]

\[x^{2} + 4x + 4 - 3 \cdot \left( \frac{16}{x} \right)^{2} =\]

\[= 2 \cdot 15 + 4 \cdot \frac{16}{x}\]

\[x^{2} + 4x + 4 - \frac{768}{x^{2}} - 32 - \frac{64}{x} =\]

\[= 0\ \ \ \ \ \ | \cdot x^{2}\]

\[x^{4} + 4x^{3} - 28x^{2} - 64x - 768 =\]

\[= 0\]

\[\left( x(x + 2) \right)^{2} - 32 \cdot x \cdot (x + 2) - 768 =\]

\[= 0\]

\[x(x + 2) = u\]

\[u^{2} - 32u - 768 = 0\]

\[D = 1024 + 3072 = 4096 = 64^{2}\]

\[u_{1,2} = \frac{32 \pm 64}{2} = - 16;\ \ 48.\]

\[x_{1}\left( x_{1} + 2 \right) = - 16;\ \ \ \ \]

\[\ x_{1}^{2} + 2x_{1} + 16 = 0 \Longrightarrow \varnothing\]

\[x_{2}\left( x_{2} + 2 \right) = 48\]

\[x_{2}^{2} + 2x_{2} - 48 = 0\]

\[D = 4 + 4 \cdot 48 = 196 = 14^{2}\]

\[x_{1} = \frac{- 2 + 14}{2} = 6;\ \ \ \ \]

\[\text{\ \ \ }x_{2} = \frac{- 2 - 14}{2} = - 8\]

\[y_{1} = \frac{16}{6} = \frac{8}{3} = 2\frac{2}{3};\ \ \ \ \ \ \ \]

\[y_{2} = \frac{16}{- 8} = - 2.\]

\(Ответ:( - 8; - 2);\ \ \left( 6;2\frac{2}{3} \right).\)