\(\Leftrightarrow\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)^2x=4\\\left(1+\dfrac{12}{y+3x}\right)^2y=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(1-\dfrac{12}{y+3x}\right)^2=\dfrac{4}{x}\left(1\right)\\\left(1+\dfrac{12}{y+3x}\right)^2=\dfrac{36}{y}\left(2\right)\end{matrix}\right.\)
\(\)\((do:x;y=0\) \(không\) \(là\) \(nghiệm\) \(của\) \(hpt\)\()\)\(\left(dk:x;y\ge0\right)\)
\(\Rightarrow\left(1+\dfrac{12}{y+3x}+1-\dfrac{12}{y+3x}\right)\left(1+\dfrac{12}{y+3x}-1+\dfrac{12}{y+3x}\right)=\dfrac{36}{y}-\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{48}{y+3x}=\dfrac{36}{y}-\dfrac{4}{x}=\dfrac{36x-4y}{xy}\)
\(\Leftrightarrow48xy=\left(36x-4y\right)\left(y+3x\right)\)
\(\Leftrightarrow48xy=36xy+108x^2-4y^2-12xy\Leftrightarrow108x^2-24xy-4y^2=0\)
\(\Leftrightarrow108\left(\dfrac{x}{y}\right)^2-24\left(\dfrac{x}{y}\right)-4=0\)
\(đặt:\dfrac{x}{y}=a\Rightarrow108a^2-24a-4=0\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{3}\Rightarrow y=3x\\a=-\dfrac{1}{9}\Rightarrow y=-9x\end{matrix}\right.\)
\(với:y=3x\Rightarrow\)\(\left(2\right)\Leftrightarrow\left(1+\dfrac{12}{6x}\right)^2=\dfrac{36}{3x}\)
đến đây đơn giản rồi giải bình thường\(\Rightarrow x=2\left(2+\sqrt{3}\right)\Rightarrow y=3x=6\left(2+\sqrt{3}\right)\)(thỏa)
\(với:y=-9x\Rightarrow\left(2\right)\Leftrightarrow\left(1+\dfrac{12}{-6x}\right)^2=\dfrac{36}{-9x}\)
giải pt =>vô nghiệm
\(\Rightarrow S=\left\{2\left(2+\sqrt{3}\right);6\left(2+\sqrt{3}\right)\right\}\)
