ĐKXĐ: \(x\ne0\)
Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=a\Rightarrow\dfrac{x^2}{4}+\dfrac{9}{x^2}-3=a^2\Rightarrow\dfrac{x^2}{4}+\dfrac{9}{x^2}=a^2+3\Rightarrow\dfrac{x^2}{2}+\dfrac{18}{x^2}=2a^2+6\)
Pt đã cho trở thành:
\(2a^2+6=13a\Leftrightarrow2a^2-13a+6=0\Rightarrow\left[{}\begin{matrix}a=6\\a=\dfrac{1}{2}\end{matrix}\right.\)
TH1: \(a=6\Rightarrow\dfrac{x}{2}-\dfrac{3}{x}=6\Leftrightarrow\dfrac{x^2-6}{2x}=6\Leftrightarrow x^2-12x-6=0\)
\(\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{42}\\x=6+\sqrt{42}\end{matrix}\right.\)
TH2: \(a=\dfrac{1}{2}\Rightarrow\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\Leftrightarrow\dfrac{x^2-6}{2x}=\dfrac{1}{2}\Leftrightarrow x^2-x-6=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy pt đã cho có 4 nghiệm
