$ĐKXĐ : x \neq 0$
Ta có : $x^2+\dfrac{1}{x^2}-\dfrac{9}{2}.(x+\dfrac{1}{x})+7 = 0 $
$⇔ (x^2+\dfrac{1}{x^2}+2)-\dfrac{9}{2}.(x+\dfrac{1}{x}) + 5 =0$
$⇔ (x+\dfrac{1}{x})^2-\dfrac{9}{2}.(x+\dfrac{1}{x}+5=0$ (1)
Đặt $x+\dfrac{1}{x} = a$ Khi đó pt $(1)$ đã cho trở thành :
$a^2-\dfrac{9}{2}a+5=0$
$⇔2a^2-9a+10=0$
$⇔2a^2-4a-5a+10=0$
$⇔2a.(a-2)-5.(a-2)=0$
$⇔(a-2).(2a-5)=0$
$⇔ \left[ \begin{array}{l}a=2\\a=\dfrac{5}{2}\end{array} \right.$
Với $a=2 ⇒ x+\dfrac{1}{x} = 2 $
$⇒ x^2-2x+1=0$
$⇔(x-1)^2=0$
$⇔x=1$ ( Thỏa mãn )
Với $a=\dfrac{5}{2} $
$⇒x+\dfrac{1}{x} = \dfrac{5}{2}$
$⇒2x^2-5x+2=0$
$⇔2x^2-x-4x+2=0$
$⇔x.(2x-1)-2.(2x-1)=0$
$⇔(2x-1).(x-2)=0$
$⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=2\end{array} \right.$ ( Thỏa mãn )
Vậy $S = ${$1,2,\dfrac{1}{2}$}
