vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).

vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).

vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).

vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).

vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).

vì \(1 + 4 = 2 + 3\),ta đặt  \(t = \left(\right. x - 1 \left.\right) \left(\right. x - 4 \left.\right) = x^{2} - 5 x + 4\) thì 

\(\left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) = x^{2} - 5 x + 6 = t + 2\)

từ đó \(\left(\right. x - 1 \left.\right) \left(\right. x - 2 \left.\right) \left(\right. x - 3 \left.\right) \left(\right. x - 4 \left.\right) + 1\)

\(= t \left(\right. t + 2 \left.\right) + 1 = t^{2} + 2 t + 1 = \left(\right. t + 1 \left.\right)^{2} \geq 0\)

Dẳng thức chỉ xảy ra khi \(t = - 1\)

hay \(x^{2} - 5 x + 4 = - 1\)

\(x^{2} - 5 x + 5 = 0\)

\(x = \frac{5 \pm \sqrt{5}}{2}\).