6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$.

6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$.
## Solution.

Domain of definition: $x \geq 0$.

From the condition, we have $\sqrt{x}=2-\frac{2 x+1}{x+2} \Leftrightarrow \sqrt{x}=\frac{2 x+4-2 x-1}{x+2}, \sqrt{x}=\frac{3}{x+2}$.

Squaring both sides of the equation, we get $x=\frac{3}{x^{2}+4 x+4} \Leftrightarrow$ $\Leftrightarrow \frac{x^{3}+4 x^{2}+4 x-9}{x^{2}+4 x+4}=0$, from which $x^{3}+4 x^{2}+4 x-9=0$ for $x \neq-2$.

Rewrite the equation as $\left(x^{3}-1\right)+\left(4 x^{2}-4\right)+(4 x-4)=0$, $(x-1)\left(x^{2}+x+1\right)+4(x-1)(x+1)+4(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}+5 x+9\right)=0$.

From this, $x-1=0$ or $x^{2}+5 x+9=0$. From the last two equations, we have $x_{1}=1 ;$ the second equation has no solutions, since $D<0$.

Answer: $x=1$.