33. $y=\sqrt{x}+\sqrt{x-1}$.

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33. $y=\sqrt{x}+\sqrt{x-1}$.
Solution. Let's find the domain of definition of each term separately. The common part of these domains will be the domain of definition of the function. For $\sqrt{x}$ we have $x \geqslant 0$, and for $\sqrt{x-1}$ we have $x \geqslant 1$. Then for the sum $\sqrt{x}+\sqrt{x-1}$ the domain of definition is $x \geqslant 1$ or $[1, \infty)$.