#### Answer

$x^{2}\sqrt x$

#### Work Step by Step

We know from the radical form of $a^{\frac{m}{n}}$ that $a^{\frac{m}{n}}=\sqrt[n] a^{m}$ (where the indicated roots are real numbers).
Therefore, $\sqrt[10] x^{25}=x^{\frac{25}{10}}=x^{\frac{5}{2}}=\sqrt x^{5}$.
$\sqrt x^{5}=\sqrt (x^{4}\times x)=x^{2}\sqrt x$