Ta có:

(8x – 7)(8x – 5)(2x – 1)(4x – 1) = 9

⇔ (8x – 7)(8x – 5)(8x – 4)(8x – 2) = 72

Đặt 8x – 5 = a

Khi đó ta có:

(a – 2)a(a + 1)(a + 3) = 72

⇔ (a² – 2a)(a² + 4a + 3) – 72 = 0

⇔ a⁴ – 4a³ + 3a² – 2a³ – 8a² – 6a – 72 = 0

⇔ a⁴ + 4a³ – 2a³ – 8a² + 3a² + 12a – 18a – 72 = 0

⇔ a³(a + 4) – 2a²(a + 4) + 3a(a + 4) – 18(a + 4) = 0

⇔ (a + 4)(a³ – 2a² + 3a – 18) = 0

⇔ (a + 4)(a³ – 3a² + a² – 3a + 6a – 18) = 0

⇔ (a + 4)[a²(a – 3) + a(a – 3) + 6(a – 3)] = 0

⇔ (a + 4)(a – 3)(a² + a + 6) = 0                              (*)

Vì \[{{\rm{a}}^2} + a + 6 = {a^2} + 2.a.\frac{1}{2} + \frac{1}{4} + \frac{{23}}{4} = {\left( {a + \frac{1}{2}} \right)^2} + \frac{{23}}{4} > 0\]

Nên \(\left( * \right) \Leftrightarrow \left[ \begin{array}{l}a + 4 = 0\\a - 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}a = - 4\\a = 3\end{array} \right.\)

Suy ra \(\left[ \begin{array}{l}8{\rm{x}} - 5 = - 4\\8{\rm{x}} - 5 = 3\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}8{\rm{x}} = 1\\8{\rm{x}} = 8\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{1}{8}\\x = 1\end{array} \right.\)

Vậy x = 1 hoặc \[{\rm{x}} = \frac{1}{8}\].