$\begin{array}{l}e)\,\,(3x-1)^2 - (x+5)^2 = 0\\ \Leftrightarrow [3x - 1 - (x+5)][3x - 1 + (x+5)] = 0\\ \Leftrightarrow (2x - 6)(4x + 4) = 0\\ \Leftrightarrow (x - 3)(x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}x - 3 = 0\\x + 1 = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =3\\x =-1\end{array}\right.\\ f)\,\,(2x-1)^2 - (x-3)^2 = 0\\ \Leftrightarrow [2x - 1 - (x-3)][2x - 1 + (x-3)] = 0\\ \Leftrightarrow (x + 2)(3x - 4) = 0\\ \Leftrightarrow \left[\begin{array}{l}x +2 = 0\\3x - 4 = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =-2\\x = \dfrac{4}{3}\end{array}\right.\\ g)\,\,(2x - 1)^2 - (4x^2 - 1) = 0\\ \Leftrightarrow (2x - 1)^2 - (2x - 1)(2x + 1) = 0\\ \Leftrightarrow (2x - 1)[2x - 1 - (2x + 1)] = 0\\ \Leftrightarrow (2x - 1).(-2) = 0\\ \Leftrightarrow 2x - 1 = 0\\ \Leftrightarrow x = \dfrac{1}{2}\\ h)\,\,x^2(x^2 + 4) - x^2 - 4 =0\\ \Leftrightarrow x^2(x^2 + 4) - (x^2 + 4) = 0\\ \Leftrightarrow (x^2 + 4)(x^2 - 1) =0\\ \Leftrightarrow (x^2 + 4)(x-1)(x+1) = 0\\ \Leftrightarrow \left[\begin{array}{l}x^2 +4 = 0\quad \text{(vô nghiệm)}\\x - 1 = 0\\x + 1 =0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =1\\x =-1\end{array}\right.\\ i)\,\,x^4 - x^3 + x^2 - x = 0\\ \Leftrightarrow x^3(x - 1) + x(x-1) = 0\\ \Leftrightarrow (x-1)(x^3 + x) = 0\\ \Leftrightarrow x(x-1)(x^2 + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x - 1 = 0\\x^2 + 1 = 0 \quad \text{(vô nghiệm)}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x =1\end{array}\right.\\ k)\,\,4x^2 - 25 - (2x - 5)(2x + 7) = 0\\ \Leftrightarrow (2x - 5)(2x + 5) - (2x - 5)(2x + 7) = 0\\ \Leftrightarrow (2x - 5)[2x + 5 - (2x + 7)] = 0\\ \Leftrightarrow (2x - 5).(-2) = 0\\ \Leftrightarrow 2x - 5 = 0\\ \Leftrightarrow x = \dfrac{5}{2}\\ l)\,\,x^3 - 8 - (x-2)(x-12) = 0\\ \Leftrightarrow (x-2)(x^2 + 2x + 4) - (x-2)(x-12) = 0\\ \Leftrightarrow (x-2)[x^2 + 2x + 4 - (x-12)] = 0\\ \Leftrightarrow (x-2)(x^2 + x + 16) = 0\\ \Leftrightarrow \Leftrightarrow \left[\begin{array}{l}x - 2 = 0\\x^2 + x + 16 = 0\quad \text{(vô nghiệm)}\end{array}\right.\\ m)\,\,2(x+3) - x^2 - 3x = 0\\ \Leftrightarrow 2(x+3) - x(x + 3) = 0\\ \Leftrightarrow (x+3)(2 - x) = 0\\ \Leftrightarrow \left[\begin{array}{l}x + 3 = 0\\2 - x = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =- 3\\x = 2\end{array}\right.\\ \end{array}$