10.1. Find all solutions of the equation: $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$.
Answer: $\frac{1}{2} \leq x \leq 1$.

Solution. First, as required, we find the domain of the variable $x$. From the non-negativity of the expressions under the square roots, we have: $2 x-1 \geq 0, x \geq \frac{1}{2}$ and $x-\sqrt{2 x-1} \geq 0, x \geq 0, x \geq \frac{1}{2}$, the second is equivalent to $(x-1)^{2} \geq 0$. Therefore, $x \geq \frac{1}{2}$.

Given the non-negativity of the left and right sides of the equation, we can square everything, obtaining $2 x+2 \sqrt{(x-1)^{2}}=2$, that is, $x+|x-1|=1$. For $x \geq 1$, we get the only solution $x=1$, and for $\frac{1}{2} \leq x<1$, we get the true identity $1=1$, so the entire interval is a solution.

Grading criteria. Correctly found domain: 1 point. Non-negativity of both sides of the equation not mentioned when squaring: deduct 1 point. Extraneous solutions: final score no higher than 3 points. Loss of one value from the set of solutions - deduct 2 points, loss of more than one - deduct 3 points.