\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\)
\(\Leftrightarrow\left[\left(x-5\right)\left(x-6\right)\right]\cdot\left[\left(x-3\right)\left(x-10\right)\right]=24x^2\)
\(\Leftrightarrow\left(x^2-11x+30\right)\left(x^2-13x+30\right)-24x^2=0\)
Đặt: \(x^2-13x+30=t\)
Lúc này PT trở thành:
\(t\left(t+2x\right)-24x^2=0\)
\(\Leftrightarrow t^2+2tx-24x^2=0\)
\(\Leftrightarrow t^2+6tx-4tx-24x^2=0\)
\(\Leftrightarrow t\left(t+6x\right)-4x\left(t+6x\right)=0\)
\(\Leftrightarrow\left(t+6x\right)\left(t-4x\right)=0\)
\(\Leftrightarrow\left(x^2-7x+30\right)\left(x^2-17x+30\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-7x+30=0\\x^2-17x+30=0\end{matrix}\right.\)
Ta có: \(x^2-7x+30=\left(x-\dfrac{7}{2}\right)^2+\dfrac{71}{4}>0\)(vô nghiệm)
=> \(x^2-17x+30=0\)
\(\Leftrightarrow\) \(\left(x-15\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-15=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)
Vậy x = 2 hoặc x = 15
