để bt có nghĩa thì:
\(\sqrt{2}\cdot x-x^2\ne0\)
\(\Leftrightarrow-x\left(x-\sqrt{2}\right)\)≠ 0
\(\Leftrightarrow\left[{}\begin{matrix}-x\ne0\\x-\sqrt{2}\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\ne\sqrt{2}\end{matrix}\right.\)
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