[ Irrational equations ]

Solve the equation $\frac{x^{3}}{\sqrt{4-x^{2}}}+x^{2}-4=0$.

#
Answer: $x=\sqrt{2}$. From the given equation, it follows that

$$
x^{3}=\left(4-x^{2}\right) \sqrt{4-x^{2}}=\left(4-x^{2}\right)^{3 / 2}
$$

By squaring both sides and taking the cube root, we get: $4-x^{2}=x^{2}$, from which $x^{2}=2$ and $x= \pm \sqrt{2}$.

Checking shows that the negative root is extraneous. Therefore, $\mathrm{x}=\sqrt{2}$ is the root of the given equation.