10.1. Solve 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, we find the domain of the equation, which is $x \geq \frac{1}{2}$ from the inner radical. The condition $x \geq \sqrt{2 x-1}$ is also satisfied, as both sides are non-negative and after squaring, it is equivalent to the inequality $(x-1)^{2} \geq 0$.

Both sides of the original equation are also non-negative, so upon squaring, we obtain the equivalent equation $2 x+2 \sqrt{(x-1)^{2}}=2 x+2|x-1|=2$, which means $|x-1|=1-x$, equivalent to $x \leq 1$. Intersecting with the domain, we get the answer $\frac{1}{2} \leq x \leq 1$.

Grading criteria. The condition $x \geq \sqrt{2 x-1}$ is not checked when finding the domain of the equation: deduct 1 point. No reference to the non-negativity of the parts of the equation when squaring: deduct 1 point.