$2.19 \frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$.
2.19 To solve the equation, we will use the following statement: if the equality $\frac{a}{b}=\frac{c}{d}$ is true, then the equality (check it!) $\frac{a+b}{a-b}=\frac{c+d}{c-d}$ is also valid. We have

$$
\begin{aligned}
& \frac{(\sqrt{x}+\sqrt[3]{x})+(\sqrt{x}-\sqrt[3]{x})}{(\sqrt{x}+\sqrt[3]{x})-(\sqrt{x}-\sqrt[3]{x})}=\frac{3+1}{3-1} ; \frac{2 \sqrt{x}}{2 \sqrt[3]{x}}=2 \\
& \sqrt{x}=2 \sqrt[3]{x} ; x^{3}=64 x^{2}
\end{aligned}
$$

Since $x \neq 0$, then $x=64$.

Answer: $x=64$.