Đặt :
$x^2 +2x + 2 = t$. Ta có $x^2 +2x +2 = (x^2+2x+1)+1 = (x+1)^2 +1$
Vì $ (x+1)^2 \ge 0 \to (x+1)^2 +1 > 0 \to t > 0$ . Ta có phương trình tương đương
$ \dfrac{t-1}{t} + \dfrac{t}{t+1} = \dfrac{7}{6} $
$\to \dfrac{(t-1)(t+1)}{t(t+1)} + \dfrac{t^2}{t(t+1)} = \dfrac{7}{6}$
$ \to \dfrac{2t^2-1}{t(t+1)} = \dfrac{7}{6}$
$ \to 6(2t^2-1) = 7t(t+1) \to 12t^2 -6 - 7t^2 -7t = 0$
$\to 5t^2 - 7t -6 = 0 \to (5t+3)(t-2) = 0$
$ \to$ \(\left[ \begin{array}{l}5t+3=0\\t-2=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}t = \dfrac{-3}{5}\ \text{(không thỏa mãn)}\\t=2\ \text{(thỏa mãn)} \end{array} \right.\)
$ t= 2 \to x^2+2x+2 = 2 \to x^2 +2x = 0 \to x(x+2)=0$
$ \to$ \(\left[ \begin{array}{l}x=0\\x+2=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
Vậy $ x \in \{0;-2 \}$
