(x^2-5x+7)^2-(2x-5)^2=0

$(x² -5x +7)² -(2x -5)² = 0$

$⇔ (x² -5x +7 +2x -5).(x² -5x +7 -2x +5) = 0$

$⇔ (x² -3x +2).(x² -7x +12) = 0$

$⇔ (x -1).(x -2).(x -3).(x -4) = 0$

$⇔ (x² -5x +4).(x² -5x +6) = 0$

$Đặt$ $t = x² -5x +5, ta$ $có:$

$(t -1).(t +1) = 0$

$⇔ t² -1 = 0$

$⇔ t² = 1$

$⇔ t = ±1$

$Thay$ $t = x² -5x +5, ta$ $có:$

$TH1:$ $x² -5x +5 = 1$

$⇔ x² -5x +4 = 0$

$⇔ (x - \frac{5}{2})² - \frac{9}{4} = 0$

$⇔ (x - \frac{5}{2} - \frac{3}{2}).(x - \frac{5}{2} + \frac{3}{2}) = 0$

$⇔ \left[ \begin{array}{l}x - \frac{5}{2} - \frac{3}{2}=0\\x - \frac{5}{2} + \frac{3}{2}=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=4\\x=1\end{array} \right.$

$TH2:$ $x² -5x +5 = -1$

$⇔ x² -5x +6 = 0$

$⇔ x² -2x -3x +6 = 0$

$⇔ x.(x -2) -3.(x -2) = 0$

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

$Vậy$ $S =$ {$4; 1; 2; 3$}