(x + 2)(x + 3)(x + 4)(x + 5) - 24 = 02
\(\Leftrightarrow\) [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24 = 0
\(\Leftrightarrow\) (x2 + 7x + 10)(x2 + 7x + 12) - 24 = 0
Đặt x2 + 7x + 11 = a. Thay vào ta có :
(a - 1)(a + 1) - 24 = 0
\(\Leftrightarrow\) a2 - 25 = 0
\(\Leftrightarrow\) (a - 5)(a + 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}a-5=0\\a+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
TH1 : a = 5
\(\Leftrightarrow\) x2 + 7x + 11 = 5
\(\Leftrightarrow\) x2 + 7x + 6 = 0
\(\Leftrightarrow\) (x + 1)(x + 6) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
TH2 : x2 + 7x + 11 = -5
\(\Leftrightarrow\) x2 + 7x + 16 = 0
\(\Leftrightarrow\) (x2 + 2.\(\frac{7}{2}\).x + \(\frac{49}{4}\)) + \(\frac{15}{4}\) = 0
\(\Leftrightarrow\) (x + \(\frac{7}{2}\))2 + \(\frac{15}{4}\) = 0 mà \(\left(x+\frac{7}{2}\right)^2\ge0\) \(\Rightarrow\) Vô lý
Vậy S = \(\left\{-1,-6\right\}\)
