Solve the following equations for $x$, verify the correctness of the obtained roots, and discuss the roots for different values of the parameters:
a) $\frac{x+a}{x-a}+\frac{x+b}{x-b}=2$,
b) $\quad c\left(\frac{d}{a b}-\frac{a b}{x}\right)+d=\frac{c^{2}}{x}$.
a) Excluding the values $x=a, x=b$, multiplying both sides of the equation by $(x-a)(x-b)$

$$
(x+a)(x-b)+(x-a)(x+b)=2(x-a)(x-b)
$$

which is

$$
2 x(a+b)=4 a b
$$

from which

$$
x=\frac{2 a b}{a+b}
$$

This only makes sense if $a+b \neq 0$.

Substitute this value of $x$ into the right side of the original equation and expand the fractions by $(a+b)$ (assuming $a+b \neq 0$)

$$
\begin{gathered}
\frac{2 a b+a^{2}+a b}{2 a b-a^{2}-a b}+\frac{2 a b+a b+b^{2}}{2 a b-a b-b^{2}}=\frac{a+3 b}{b-a}+\frac{3 a+b}{a-b}=\frac{-a-3 b+3 a+b}{a-b}= \\
=\frac{2(a-b)}{a-b}=2
\end{gathered}
$$

assuming $a-b \neq 0$.

If $a-b=0$, or $a=b$, then our equation is

$$
\frac{x+a}{x-a}=1
$$

which is a contradiction if $a(=b) \neq 0$; and an identity if $a(=b)=0$. The assumption $a=0, b \neq 0$ (or vice versa) also leads to a contradiction: $\frac{x+b}{x-b}=1$.

Thus, the solution to our equation is $x=\frac{2 a b}{a+b}$, provided that $a+b \neq 0, a-b \neq 0, a \neq 0$, and $b \neq 0$, meaning the parameters are non-zero quantities, and their absolute values are also different.

b) Excluding the values $a=0, b=0, x=0$, multiplying both sides by $a b x$

$$
c d x-a^{2} b^{2} c+a b d x=a b c^{2}
$$

from which

$$
x=\frac{a b c(a b+c)}{d(a b+c)}=\frac{a b c}{d}
$$

assuming $a b+c \neq 0, d \neq 0$ and (since $x \neq 0$) $c \neq 0$.

If $a b+c=0$, or $a b=-c$, then our equation is:

$$
-d+\frac{c^{2}}{x}+d=\frac{c^{2}}{x}
$$

an identity.

Otherwise, if all four parameters are non-zero, $x=\frac{a b c}{d}$ is indeed a solution, as we can verify by substitution.

Right side:

$$
\frac{c d}{a b}-\frac{a b c d}{a b c}+d=\frac{c d}{a b}
$$

Left side:

$$
\frac{c^{2} d}{a b c}=\frac{c d}{a b}
$$

Zsombok Zoltán (Bp., IV., Konyves Kalman g. II. o. t.)