6.154. $\frac{x^{2}+2 x+1}{x^{2}+2 x+2}+\frac{x^{2}+2 x+2}{x^{2}+2 x+3}=\frac{7}{6}$.
## Solution.

Let $x^{2}+2 x=y$, then the equation in terms of $y$ becomes $\frac{y+1}{y+2}+\frac{y+2}{y+3}-\frac{7}{6}=0 \Leftrightarrow \frac{6(y+1)(y+3)+6(y+2)^{2}-7(y+2)(y+3)}{6(y+2)(y+3)}=0 \Leftrightarrow$
$\Leftrightarrow \frac{5 y^{2}+13 y}{(y+2)(y+3)}=0,5 y^{2}+13 y=0$ for $y \neq-2$ and $y \neq-3 ; y(5 y+13)=0$, from which $y_{1}=0, y_{2}=-\frac{13}{5}$. Then $x^{2}+2 x=0$ or $x(x+2)=0 \Rightarrow x_{1}=0$, $x_{2}=-2 ; x^{2}+2 x=-\frac{13}{5}, 5 x^{2}+10 x+13=0$, no solutions, since $D<0$.

Answer: $x_{1}=0, x_{2}=-2$.