388. Solve the equations:
a) $\frac{x^{3}}{\sqrt{4-x^{2}}}+x^{2}-4=0$
b) $\sqrt{17 x^{2}+7 x+0.5}=13 x^{2}+5 x+0.5$.
$\triangle$ a) Transform the equation:

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

Extract the cubic root from both sides of the last equation (thus obtaining an equivalent equation):

$$
x=\left(4-x^{2}\right)^{\frac{1}{2}}, \quad x=\sqrt{4-x^{2}}
$$

Square both sides of the new equation:

$$
x^{2}=4-x^{2}, \quad x^{2}=2 ; \quad x= \pm \sqrt{2}
$$

It is necessary to check the found roots. It is more convenient to check them using the equation $x=\sqrt{4-x^{2}}$. The value $x=\sqrt{2}$ satisfies this equation, while the value $x=-\sqrt{2}$ does not.

b) Square both sides of the equation and simplify the resulting algebraic equation by collecting all terms on the left side:

$17 x^{2}+7 x+0.5=169 x^{4}+25 x^{2}+0.25+130 x^{3}+13 x^{2}+5 x$, $169 x^{4}+130 x^{3}+21 x^{2}-2 x-0.25=0$, $676 x^{4}+520 x^{3}+84 x^{2}-8 x-1=0$.

The obtained equation does not have integer roots, as the divisors of the free term, equal to $\pm 1$, are not its roots. Let's try to factor the polynomial on the left side of the equation into quadratic trinomials with integer coefficients:

$$
676 x^{4}+520 x^{3}+84 x^{2}-8 x-1=\left(a x^{2}+b x+1\right)\left(c x^{2}+d x-1\right)
$$

where $a, b, c, d$ are integers. Expand the brackets on the right side of this identity and equate the coefficients of like terms on the left and right sides:

$$
\left\{\begin{array}{l}
a c=676 \\
a d+b c=520 \\
c-a+b d=84 \\
d-b=-8
\end{array}\right.
$$

Does this system of equations have a solution in integers? Note the first equation $a c=676$. The numbers $a$ and $c$ are divisors of the number 676. Suppose the number $a$ is even. Then from the second equation, the number $b c$ is even. If at the same time the number $b$ is even, then from the third equation $c$ is even. It turns out that the numbers $a$ and $c$ are both even.

The last fact simplifies solving the system by enumeration. Since $676=26^{2}=2^{2} \cdot 13^{2}$, the possible values for $a$ and $c$ are either $\pm 26$ and $\pm 26$, or $\pm 2$ and $\pm 338$ (with the same signs for $a$ and $c$).

Let's take, for example, $a=c=26$. Then

$$
268+26 d=520 \text { (i.e. } b+d=20 \text { ), } b d=84, \quad d-b=-8 \text {. }
$$

The last system has the solution $b=14, d=6$. Therefore, the fourth-degree equation takes the following form:

$$
\left(26 x^{2}+14 x+1\right)\left(26 x^{2}+6 x-1\right)=0
$$

Its roots are -

$$
x_{1,2}=\frac{-7 \pm \sqrt{23}}{26}, \quad x_{3,4}=\frac{-3 \pm \sqrt{35}}{26} .
$$

Checking these roots in the original equation is not necessary in this case (think about why).

Answer: a) $\sqrt{2}$; b) $(-7 \pm \sqrt{23}) / 26,(-3 \pm \sqrt{35}) / 26$.