Hệ PT hay!

Đề bài: (Câu hỏi của bạn Huyền Trang Nguyễn hỏi trên facebook Trợ Giúp Toán Học)
Giải hệ phương trình sau: $\left\{ \begin{array}{l} x\sqrt {{y^2} + 6}  + y\sqrt {{x^2} + 3}  = 7xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\ y\sqrt {{y^2} + 6}  + x\sqrt {{x^2} + 3}  = {x^2} + {y^2} + 2\,\,\,\,\left( 2 \right) \end{array} \right.$
Giải:

$$\begin{array}{l}  \bullet \,Khi\,xy = 0 \Rightarrow \left[ \begin{array}{l} x = 0 \Rightarrow \left( 1 \right) \Leftrightarrow y\sqrt 3  = 0 \Rightarrow y = 0 \Rightarrow \left( 2 \right) \Leftrightarrow 0 = 2\left( {V.Ly'} \right)\\ y = 0 \Rightarrow \left( 1 \right) \Leftrightarrow x\sqrt 6  = 0 \Rightarrow x = 0 \Rightarrow \left( 2 \right) \Leftrightarrow 0 = 2\left( {V.Ly'} \right) \end{array} \right.\\  \bullet \,Khi\,xy \ne 0 \Rightarrow \left( 1 \right) \Leftrightarrow \sqrt {1 + \frac{3}{{{x^2}}}}  + \sqrt {1 + \frac{6}{{{y^2}}}}  = 7\& \left( 2 \right) \Leftrightarrow x\left( {\sqrt {{x^2} + 3}  - x} \right) + y\left( {\sqrt {{y^2} + 6}  - y} \right) = 2\\  \Leftrightarrow \frac{{3x}}{{\sqrt {{x^2} + 3}  + x}} + \frac{{6y}}{{\sqrt {{y^2} + 6}  + y}} = 2 \Leftrightarrow \frac{3}{{\sqrt {1 + \frac{3}{{{x^2}}}}  + 1}} + \frac{6}{{\sqrt {1 + \frac{6}{{{y^2}}}}  + 1}} = 2\\ Coi\,\left\{ \begin{array}{l} a = \sqrt {1 + \frac{3}{{{x^2}}}} \left( {a > 1} \right)\\ b = \sqrt {1 + \frac{6}{{{y^2}}}} \left( {b > 1} \right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a + b = 7\\ \frac{3}{{a + 1}} + \frac{6}{{b + 1}} = 2\\ a,b > 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = 7 - a\\ 2{a^2} - 11a + 14 = 0\\ a,b > 1 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} \left( {a;b} \right) = \left( {2;5} \right) \Rightarrow \left\{ \begin{array}{l} \frac{3}{{{x^2}}} = {a^2} - 1 = 3 \Leftrightarrow {x^2} = 1 \Leftrightarrow x =  \pm 1\\ \frac{6}{{{y^2}}} = {b^2} - 1 = 24 \Leftrightarrow {y^2} = \frac{1}{4} \Leftrightarrow y =  \pm \frac{1}{2} \end{array} \right.\\ \left( {a;b} \right) = \left( {\frac{7}{2};\frac{7}{2}} \right) \Rightarrow \left\{ \begin{array}{l} \frac{3}{{{x^2}}} = {a^2} - 1 = \frac{{15}}{4} \Leftrightarrow {x^2} = \frac{4}{5} \Leftrightarrow x =  \pm \frac{{2\sqrt 5 }}{5}\\ \frac{6}{{{y^2}}} = {b^2} - 1 = \frac{{15}}{4} \Leftrightarrow {y^2} = \frac{5}{8} \Leftrightarrow y =  \pm \frac{{\sqrt {10} }}{4} \end{array} \right. \end{array} \right.\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow S = \left\{ {\left( { \pm 1; \pm \frac{1}{2}} \right),\left( { \pm \frac{{2\sqrt 5 }}{5}; \pm \frac{{\sqrt {10} }}{4}} \right)} \right\} \end{array}$$