Tìm x biết (x – 7)(x² – 9x + 20)(x – 2) = 72.

(x − 7)(x² − 9x + 20)(x – 2) = 72

⇒ (x² − 2x − 7x + 14)(x² − 9x + 20) – 72 = 0

⇒ (x² − 9x + 14)(x² − 9x + 14 + 6) – 72 = 0

Đặt x² − 9x + 14 = a ta có :

a(a + 6) – 72 = 0

⇔ a² + 6a – 72 = 0

⇔ a² − 6a + 12a – 72 = 0

⇔ a(a − 6) + 12(a − 6) = 0

⇔(a − 6)(a + 12) = 0

⇔ $\left[\begin{array}{l}a=6\\ a=-12\end{array}\right.$

* TH1 : a = 6 

⇒ x² − 9x + 14 = 6

⇔ x² − 9x + 14 − 6 = 0

⇔ x² − 9x + 8 = 0

⇔ x² − 8x − x + 8 = 0

⇔ x(x − 8) − (x − 8) = 0

⇔ (x − 8)(x – 1) = 0

⇔  $\left[\begin{array}{l}x=8\\ x=1\end{array}\right.$

*TH2 : a = −12

⇒ x² − 9x + 14 = −12

⇔ x² − 9x + 14 + 12 = 0

⇔ x² − 9x + 26 = 0

⇔ ${\left(x-\frac{9}{2}\right)}^2$ + $\frac{23}{4}$= 0 (Vô lý)

Vậy x ∈ {1; 8}.