6.222.

$$
\left\{\begin{array}{l}
\sqrt{\frac{x+1}{x+y}}+\sqrt{\frac{x+y}{x+1}}=2 \\
\sqrt{\frac{x+1}{y+2}}-\sqrt{\frac{y+2}{x+1}}=1.5
\end{array}\right.
$$
Solution.

Domain of definition: $\left\{\begin{array}{l}\frac{x+1}{x+y}>0, \\ \frac{x+1}{y+2}>0 .\end{array}\right.$

Rewrite the system of equations as

$\left\{\begin{array}{l}\sqrt{\frac{x+1}{x+y}}+\frac{1}{\sqrt{\frac{x+1}{x+y}}}-2=0, \\ \sqrt{\frac{x+1}{y+2}}-\frac{1}{\sqrt{\frac{x+1}{y+2}}}-1.5=0 .\end{array}\right.$

Multiplying the first equation of the system by $\sqrt{\frac{x+1}{x+y}} \neq 0$, and the second by $\sqrt{\frac{x+1}{y+2}} \neq 0$, we get

$$
\left\{\begin{array} { l } 
{ ( \sqrt { \frac { x + 1 } { x + y } } ) ^ { 2 } - 2 \sqrt { \frac { x + 1 } { x + y } } + 1 = 0 , } \\
{ ( \sqrt { \frac { x + 1 } { y + 2 } } ) ^ { 2 } - 1.5 \sqrt { \frac { x + 1 } { y + 2 } } - 1 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(\sqrt{\frac{x+1}{x+y}}-1\right)^{2}=0, \\
\left(\sqrt{\frac{x+1}{y+2}}\right)^{2}-1.5 \sqrt{\frac{x+1}{y+2}}=1
\end{array} \Rightarrow\right.\right.
$$

$$
\sqrt{\frac{x+1}{x+y}}=1, \frac{x+1}{x+y}=1, \text { from which } x+1=x+y, y=1
$$

The second equation of the system is solved as a quadratic equation in terms of $t=\sqrt{\frac{x+1}{y+2}}: t^{2}-1.5 t-1=0$.

$$
\text { Hence } t_{1}=\sqrt{\frac{x+1}{y+2}}=2 ; t_{2}=\sqrt{\frac{x+1}{y}+2}=-\frac{1}{2} ; 2 \sqrt{\frac{x+1}{y+2}}=-1 \text { - has no }
$$

solutions.

$$
\text { Then } \frac{x+1}{y+2}=\frac{x+1}{1+2}=4, x+1=12 ; x=11 \text {. }
$$

Answer: $(11 ; 1)$.