$ĐKXĐ : x \neq 0$

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 = ${$1,2,\dfrac{1}{2}$}

 

$\frac{x² +1}{x²} - \frac{9}{2}.\frac{x +1}{x} +7 = 0$ $(ĐKXĐ:$ $x \neq 0)$

$⇔ \frac{x² +1}{x²} - \frac{9(x +1)}{2x} +7 = 0$

$⇔ \frac{2.(x² +1)}{2x²} - \frac{9x(x +1)}{2x²} + \frac{14x²}{2x²} = 0$

$⇒ 2.(x² +1) -9x.(x +1) +14x² = 0$

$⇔ 2x² +2 -9x² -9x +14x² = 0$

$⇔ 7x² -9x +2 = 0$

$⇔ 7x.(x -1) -2.(x -1) = 0$

$⇔ (x -1).(7x -2) = 0$

$⇔ \left[ \begin{array}{l}x -1=0\\7x -2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\frac{2}{7}\end{array} \right.$

$Vậy$ $S =$ {$2; \frac{2}{7}$}