a) x(x + 1)(x - 1)(x + 2) = 24

⇔ (x2 + x) (x2 - x + 2x + - 2) = 24

⇔ (x2 + x) (x2 + x - 2) = 24

Đặt: x2 + x -1 = t

⇔ (t - 1) (t + 1) = 24

⇔ t2 - 1 - 24 = 0

⇔ t2 - 25 = 0

⇔ (t - 5) (t + 5) = 0

⇔ (x2 + x - 1 - 5) (x2 + x - 1 + 5) = 0

⇔ (x2 + x - 6) (x2 + x + 4) = 0

⇔ (x2 + x + $\frac{1}{4}$ + $\frac{15}{4}$ ) ($x^{2}$  + 3x - 2x - 6) = 0

⇔  (( x2 + x + $\frac{1}{4}$ ) + $\frac{15}{4}$) ( ($x^{2}$  + 3x) -(2x + 6) = 0

⇔  + $\frac{15}{4}$ ) (x(x + 3) - 2(x + 3) ) = 0.

⇔ (x - 2) (x + 3) = 0

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

⇔ \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

 Vậy: S = ( 2; -3)

c. ( x + 2) (x + 3)(x - 5)(x - 6) = 180

⇔ ( x2 + 2x - 5x -10) (x2 + 3x - 6x - 18) = 180

⇔ ( x2 - 3x -10 ) ( x2 - 3x - 18) = 180

Đặt : x2 - 3x  - 14 = t

⇔ ( x + 14) (x - 14) = 180

⇔ t2 - 16 - 180 = 0

⇔ t2 - 196 = 0

⇔ (t -+ 14) (t - 14) = 0

⇔ (x2 - 3x  - 14 + 14) ( x2 - 3x  - 14  - 14) = 0

⇔ (x2 - 3x) ( x2 - 3x - 28) =0

⇔ x(x - 3) (x2 - 7x + 4x - 28) = 0

⇔ x(x - 3) ( x(x - 7) + 4(x - 7) ) 0

⇔ x ( x-3) (x - 7) (x + 4) = 0

TH1: x = 0

TH2: x - 3 = 0

⇔ x = 3

TH3: x - 7 = 0

⇔ x = 7

TH4: x + 4 = 0

⇔ x = -4

Vậy: S = (0; 3; 7; -4)

b. ( x - 4) (x - 5)(x - 6)(x-7) = 1680

⇔ ( x - 4)(x-7) (x - 5)(x - 6) = 1680

⇔ (x2 - 11x + 28) (x2 - 11x + 30) = 1680

  Đăt: x2 - 11x + 29 = t

⇒ ( t - 1) (t + 1) = 1680

⇔ t2 - 1 = 1860

⇔ t2 = 1861

⇔ t = ± 41

TH1:  t = -41

⇔ x2 - 11x + 29 = -41

⇔ x2 - 11x + 70 = 0

⇔ x2 - 2. $\frac{11}{2}$ .x  + $\frac{121}{4}$ - $\frac{121}{4}$ + 70 = 0

⇔ ( x - $\frac{11}{2}$ )$^{2}$  + $\frac{159}{4}$ = 0 ( vô nghiệm)

TH2: t = 41

⇔ x2 - 11x + 29 = 41

⇔ x2 - 11x -12 = 0

⇔ x2 + x - 12x - 12 = 0

⇔ x(x+1) - 12(x + 1) = 0

⇔ (x + 1) (x - 12) = 0

    x + 1 = 0                 x - 12 = 0

⇔  x = -1                   ⇔ x = 12

KL: S = (-1 ; 12)