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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}\]