$\frac{1}{x}$+ $\frac{1}{x+2}$= $\frac{12}{35}$ [Mẫu chung: 35x(x+2)] (Điều kiện:\(\left[ \begin{array}{l}x≠0\\x+2≠0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x≠0\\x≠-2\end{array} \right.\) )
⇔ $\frac{1.35(x+2)}{35x(x+2)}$ + $\frac{1.35x}{35x(x+2)}$ = $\frac{12.x(x+2)}{35x(x+2)}$
⇔35(x+2) + 35x = 12x(x+2)
⇔ 35x +70 + 35x= 12x²+24x
⇔ -12x² + 70x - 24x + 70=0
⇔ -12x² + 46x + 70 = 0
(a= -12; b= 46; c= 70)
Ta có Δ= b²- 4ac
= 46² - (4 . -12 . 70)
= 5476
Vì Δ > 0 nên phương trình có 2 nghiệm phân biệt
\(\left[ \begin{array}{l}x_{1}=\frac{-b +\sqrt{Δ} }{2a}= \frac{-46 +\sqrt{5476} }{2. -12} = -\frac{7}{6}\\x_{2}=\frac{-b -\sqrt{Δ} }{2a}= \frac{-46 -\sqrt{5476} }{2. -12}=5\end{array} \right.\)
Vậy S={$-\frac{7}{6}$ ; 5)
