\(pt\left(1\right)\Rightarrow x>0;\text{ }pt\left(2\right)\Rightarrow y>0\)
\(pt\left(1\right)\Leftrightarrow x^2+2=3xy^2;\text{ }pt\left(2\right)\Leftrightarrow y^2+2=3x^2y\)
\(\Rightarrow x^2+2-\left(y^2+2\right)=3xy^2-3x^2y\)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+3xy\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+3xy\right)=0\)
\(\Leftrightarrow x=y\text{ (do }x+y+3xy>0\text{)}\)
Thay .....
\(x^2+2=3x^3\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)
\(\Leftrightarrow x-1=0\text{ hoặc }3x^2+2x+2=0\text{ (vô nghiệm)}\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
Kết luận: (x;y) = (1;1).
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