\(\left(x-2\right)\left(x-4\right)\left(x-10\right)\left(x-5\right)-54x^2=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]=54x^2\\ \Leftrightarrow\left(x^2-12x+20\right)\left(x^2-9x+20\right)=54x^2\)

Với x=0: ko phải nghiệm của pt

Với x≠0

\(pt\Leftrightarrow\dfrac{x^2-12x+20}{x}.\dfrac{x^2-9x+20}{x}=\dfrac{54x^2}{x^2}\\ \Leftrightarrow\left(x-12+\dfrac{20}{x}\right)\left(x-9+\dfrac{20}{x}\right)=54\left(1\right)\)

Đặt \(x+\dfrac{20}{x}=y\)

\(\left(1\right)\Leftrightarrow\left(y-12\right)\left(y-9\right)=54\\ \Leftrightarrow y^2-21y+108-54=0\\ \Leftrightarrow y^2-21y+54=0\\ \Leftrightarrow\left[{}\begin{matrix}y=3\\y=18\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{20}{x}-3=0\\x+\dfrac{20}{x}-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+20=0\left(vô.lí\right)\\x^2-18x+20=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9+\sqrt{61}\\x=9-\sqrt{61}\end{matrix}\right.\)

\(\left(x-2\right)\left(x-10\right)\left(x-4\right)\left(x-5\right)-54x^2=0\)

\(\Leftrightarrow\left(x^2+20-12x\right)\left(x^2+20-9x\right)-54x^2=0\)

\(\Leftrightarrow\left(x^2+20\right)^2-21x\left(x^2+20\right)+108x^2-54x^2=0\)

\(\Leftrightarrow\left(x^2+20\right)^2-21x\left(x^2+20\right)+54x^2=0\)

\(\Leftrightarrow\left(x^2+20\right)^2-3x\left(x^2+20\right)-18x\left(x^2+20\right)+54x^2=0\)

\(\Leftrightarrow\left(x^2+20\right)\left(x^2+20-3x\right)-18x\left(x^2+20-3x\right)=0\)

\(\Leftrightarrow\left(x^2+20-3x\right)\left(x^2+20-18x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+20=0\left(vô-nghiệm\right)\\x^2-18x+20=0\end{matrix}\right.\)

\(\Rightarrow x^2-18x+81-61=0\)

\(\Rightarrow\left(x-9\right)^2=61\)

\(\Rightarrow x-9=\pm\sqrt{61}\)

\(\Rightarrow x=9\pm\sqrt{61}\)

Đặt \(A=\left(x-2\right)\left(x-4\right)\left(x-5\right)\left(x-10\right)-54x^2\)

\(=\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]-54x^2\)

\(=\left(x^2-12x+20\right)\left(x^2-9x+20\right)-54x^2\)

Đặt \(x^2-12x+20=t\)

Khi đó: \(A=t\left(t+3x\right)-54x^2\)

\(=t^2+3tx-54x^2\)

\(=t\left(t-6x\right)+9x\left(t-6x\right)\)

\(=\left(t-6x\right)\left(t+9x\right)\)

\(=\left(x^2-18x+20\right)\left(x^2-3x+20\right)\)