\(x^2-x-18+\dfrac{72}{x^2-x}=0\) ( ĐK : \(x\ne0\) và \(x\ne1\) )
\(\Leftrightarrow x^2-x+\dfrac{72}{x^2-x}=18\)
Đặt \(x^2-x=a\) . Phương trình trở thành :
\(a+\dfrac{72}{a}=18\)
\(\Leftrightarrow a^2-18a+72=0\)
\(\Leftrightarrow\left(a-6\right)\left(a-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-6=0\\a-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=6\\a=12\end{matrix}\right.\)
Với \(a=6\) :
\(\Leftrightarrow x^2-x=6\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Với \(a=12\) :
\(\Leftrightarrow x^2-x=12\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{-2;-3;3;4\right\}\)
