Sure, here is the translated text:

```
Solve the following equation and verify the solution by substitution:

\[
\begin{aligned}
& \frac{x(x+3)-4\left\{3-[5 x-(1-2 x)] \frac{(x-1)^{2}}{4}\right\}}{-\frac{4 x}{3}-\left\{-1-\left[\frac{5}{2}\left(x+\frac{6}{5}\right)-x\right]+\frac{x}{6}\right\}}= \\
= & \frac{x}{2}(x+6)-\{-x[-3(x-4)+2(3 x-5)-10]\} .
\end{aligned}
\]
```
Let's denote the numerator on the left side of the equation as $A$, the denominator as $B$, and the expression on the right side as $C$.

$$
\begin{gathered}
A=x^{2}+3 x-12+20 x-4+8 x+x^{2}-2 x+1=2 x^{2}+29 x-15 \\
B=-\frac{4}{3} x+1+\frac{5}{2} x+3-x-\frac{1}{6} x=4 \\
C=\frac{x^{2}}{2}+3 x-3 x^{2}+12 x+6 x^{2}-10 x-10 x=\frac{7}{2} x^{2}-5 x
\end{gathered}
$$

Thus, we have simplified our equation to the following form:

$$
\begin{gathered}
\frac{2 x^{2}+29 x-15}{4}=\frac{7}{2} x^{2}-5 x \\
2 x^{2}+29 x-15=14 x^{2}-20 x \\
12 x^{2}-49 x+15=0
\end{gathered}
$$

from which

$$
x_{1}=\frac{49+41}{24}=\frac{90}{24}=\frac{15}{4}, \quad x_{2}=\frac{49-41}{24}=\frac{8}{24}=\frac{1}{3}
$$

We have only performed transformations that preserved the equivalence of the equations, so the roots we obtained must satisfy the original equation. We can verify this by substitution.

Substitute $\frac{15}{4}$ for $x$:

$$
\begin{gathered}
A=\frac{15}{4}\left(\frac{15}{4}+3\right)-4\left\{3-\left[\frac{75}{4}-\left(1-\frac{30}{4}\right)\right]-\frac{\left(\frac{15}{4}-1\right)^{2}}{4}\right\}=\frac{15}{4} \cdot \frac{27}{4}- \\
-4\left\{3-\left[\frac{75}{4}+\frac{26}{4}\right]-\frac{1}{4}\left(\frac{11}{4}\right)^{2}\right\}=\frac{405}{16}-12+101+\frac{121}{16}=\frac{263}{8}+89= \\
=\frac{263+712}{8}=\frac{975}{8}
\end{gathered}
$$

$B=4$ for all $x$. Therefore, the left side is $\frac{A}{B}=\frac{975}{32}$.

The right side of the equation is

$$
\begin{aligned}
& C=\frac{15}{8}\left(\frac{15}{4}+6\right)\left\{-\frac{15}{4}\left[-3\left(\frac{15}{4}-4\right)+2\left(3 \cdot \frac{15}{4}-5\right)-10\right]\right\}= \\
& =\frac{15}{8} \cdot \frac{39}{4}+\frac{15}{4}\left[\frac{3}{4}+\frac{50}{4}-\frac{40}{4}\right]=\frac{585}{32}+\frac{15}{4} \cdot \frac{13}{4}=\frac{585+390}{32}=\frac{975}{32}
\end{aligned}
$$

Substituting $x_{2}=\frac{1}{3}$:

$$
\begin{aligned}
A & =\frac{1}{3}\left(\frac{1}{3}+3\right)-4\left\{3-\left[\frac{5}{3}-\left(1-\frac{2}{3}\right)\right]-\frac{\left(\frac{1}{3}-1\right)^{2}}{4}\right\}= \\
& =\frac{1}{3} \cdot \frac{10}{3}-4\left\{3-\frac{4}{3}-\frac{1}{4} \cdot \frac{4}{9}\right)=\frac{10}{9}-4 \cdot \frac{14}{9}=-\frac{46}{9}
\end{aligned}
$$

and since $B=4$, the left side is $\frac{A}{B}=-\frac{23}{18}$.

$$
\begin{gathered}
C=\frac{1}{6}\left(\frac{1}{3}+6\right)-\left\{-\frac{1}{3}\left[-3\left(\frac{1}{3}-4\right)+2\left(3 \cdot \frac{1}{3}-5\right)-10\right]\right\}= \\
=\frac{1}{6} \cdot \frac{19}{3}+\frac{1}{3}[11-8-10]=\frac{19-42}{18}=-\frac{23}{18}
\end{gathered}
$$

Thus, this root also satisfies the equation.

Bartha Gyöngyi (Bp., VIII., Apáczai Csere g. I. o. t.)