Tìm x biết: (x – 2)(x – 4)(x – 5)(x – 10) – 54x² = 0.

(x – 2)(x – 4)(x – 5)(x – 10) – 54x² = 0

⇔ [(x– 2)(x – 10)][(x – 4)(x – 5)] – 54x² = 0

⇔ (x² – 12x + 20)(x² – 9x + 20) – 54x² = 0

Đặt x² – 12x + 20 = t

Khi đó ta có:

t(t + 3x) – 54x² = 0

⇔ t² + 3xt – 54x² = 0

⇔ t(t – 6x) + 9x(t – 6x) = 0

⇔ (t + 9x)(t – 6x) = 0

⇔ (x² – 18x + 20)(x² – 3x + 20) = 0

⇔ \(\left[ \begin{array}{l}{x^2} - 18x + 20 = 0\\{x^2} - 3x + 20 = 0\end{array} \right.\)

Nếu x² – 18x + 20 = 0

⇔ (x – 9)² – 61 = 0

⇔ (x – 9)² = 61

⇔ \(\left[ \begin{array}{l}x = 9 + \sqrt {61} \\x = 9 - \sqrt {61} \end{array} \right.\)

Nếu x² – 3x + 20 = 0

⇔ \[{\left( {x - \frac{3}{2}} \right)^2} + \frac{{71}}{4} \ge \frac{{71}}{4} > 0\] nên phương trình vô nghiệm.

Vậy x = \(\left\{ {9 + \sqrt {61} ;9 - \sqrt {61} } \right\}\).