10.2. Solve the equation $4 x=2+\frac{x}{\sqrt{1+x}+1}$.
Answer: $x=\frac{9}{16}$. Solution. By multiplying the fraction on the right side by the expression $(\sqrt{1+x}-1)$ (conjugate to the denominator), after canceling out by $x \neq$ Oh, we get the equation $4 x=1+\sqrt{1+x}$. Then, by canceling out the expression on the right side and transforming it in the same way as before, we will have the equation $4(\sqrt{1+x}-$ $1)=1$. From this, $1+x=\frac{25}{16}$, i.e., $x=\frac{9}{16}$. Comment. Note that in this solution, it is not necessary to check the root $x=9 / 16$ by substituting it into the original equation (although this is not difficult). The reason is that in our transformations, multiplication by the conjugate expression was performed, which is zero only at $x=0$ (which, as noted, is not a root). At the end of the solution, the equation with the square root on the left side and the positive number $5 / 4$ on the right side was squared. Due to the positivity of the right side, such squaring also gives an equivalent equation. In other solution methods encountered in the participants' works, equations where both the right and left sides contained functions of $x$ (usually linear) were squared. In such solution methods, the equations obtained are generally not equivalent (they are consequences of the original equation) and there is no guarantee that extraneous roots will not be obtained. Therefore, in this case, if participants obtained $x=9 / 16$ at the end but did not perform the check, the score was reduced.