Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=195x^2\)
=>\(\left(x^2-7x+6\right)\left(x^2-5x+6\right)=195x^2\)
=>\(\left(x^2-5x+6\right)^2-2x\left(x^2-5x+6\right)-195x^2=0\)
=>\(\left(x^2-5x+6-15x\right)\left(x^2-5x+6+13x\right)=0\)
=>\(\left(x^2-20x+6\right)\left(x^2+7x+6\right)=0\)
=>\(\left(x^2-20x+6\right)\left(x+1\right)\left(x+6\right)=0\)
=>\(\left[{}\begin{matrix}x^2-20x+6=0\\x+1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-20x+100=94\\x=-1\\x=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left(x-10\right)^2=94\\x=-1\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-10=\pm\sqrt{94}\\x=-1\\x=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\pm\sqrt{94}\\x=-1\\x=-6\end{matrix}\right.\)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=195x^2\)
\(\Leftrightarrow\left(x-1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)=195x^2\)
\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-5x+6\right)=195x^2\)
\(\Leftrightarrow\left(x-7+\dfrac{6}{x}\right)\left(x-5+\dfrac{6}{x}\right)=195\) (Chia \(2\) vế cho \(x^2\left(x\ne0\right)\))
Đặt \(t=x+\dfrac{6}{x}\)
\(PT\Leftrightarrow\left(t-7\right)\left(t-5\right)=195\)
\(\Leftrightarrow t^2-12t-160=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=20\\t=-8\end{matrix}\right.\)
\(TH_1:t=20\Rightarrow x+\dfrac{6}{x}=20\Rightarrow x^2-20x+6=0\)\(\Rightarrow x=10\pm\sqrt{94}\left(tm\right)\)
\(TH_2:t=-8\Rightarrow x+\dfrac{6}{x}=-8\Rightarrow x^2+8x+6=0\Rightarrow x=-4\pm\sqrt{10}\left(tm\right)\)
Vậy \(x\in\left\{10\pm\sqrt{94};-4\pm\sqrt{10}\right\}\) thỏa đề bài
