Solve the following equation:

$$
\frac{(a+b)(c-x)}{a^{2}}-\frac{(b+c)(x-2 c)}{b c}-\frac{(c+a)(c-2 x)}{a c}=\frac{(a+b) c}{a b}+2
$$

Examine the case when $a: b: c=6: 3: 4$.
The values of $a, b, c$ cannot be 0, otherwise the equation would not make sense. By removing the fractions, we obtain the following equivalent equation:

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

From this, through the usual rearrangement steps, we get the following equation:

$$
x\left(a^{2} b-a^{2} c+a b c-b^{2} c\right)=c\left(a^{2} b-a^{2} c+a b c-b^{2} c\right)
$$

If the coefficient of $x$ is not 0, then $x=c$.

If the coefficient of $x$ is 0, then any number is a solution to the equation. This follows from the special case to be examined; the ratio $a: b: c=6: 3: 4$ means that for some value of $u$, $a=6 u, b=3 u, c=4 u$, and substituting this into the coefficient of $x$, and the right side becomes 0, and substituting into the original equation, $x$ cancels out on the left side, and the value of both sides is 4; we arrive at an identity.