[tex]\it x+2+\dfrac{2}{y+1+\dfrac{1}{z+1}}=\dfrac{23}{5}|_{-2} \Rightarrow x+\dfrac{2}{\dfrac{(z+1)(y+1)+1}{z+1}}=\dfrac{13}{5} \Rightarrow \\ \\ \\ \Rightarrow x+\dfrac{2(z+1)}{(z+1)(y+1)+1}=2+\dfrac{3}{5} \Rightarrow x=2;\ \ \ \dfrac{2(z+1)}{(z+1)(y+1)+1}=\dfrac{3}{5} \Rightarrow\\ \\ \\ \Rightarrow 10(z+1) =3(z+1)(y+1)+3 \Rightarrow10(z+1)-3(z+1)(y+1)=3 \Rightarrow[/tex]
[tex]\it \Rightarrow (z+1)(10-3y-3)=3 = 3\cdot1 \Rightarrow \begin{cases}\ \it z+1=3 \Rightarrow z=2\\ \\ \it 7-3y=1 \Rightarrow y=2\end{cases}[/tex]
Prin urmare, x = y = z = 2
