Let's simplify the following expressions:

$$
\frac{\frac{2 x+1}{x+1}-\frac{2 x-1}{x}}{\frac{2 x-1}{x-1}-\frac{2 x+1}{x}}
$$

$$
\frac{x^{2}+x}{x^{2}+3 x+2}-\frac{x^{2}-3 x+2}{x^{2}-x}+\frac{x^{2}-7 x+12}{x^{2}-5 x+6}
$$
I. The numerator and denominator of the $K_{1}$ expression can each be written as a $2-2$ term as follows:

$$
\begin{array}{ll}
\frac{2 x+1}{x+1}=2-\frac{1}{x+1}, & \frac{2 x-1}{x}=2-\frac{1}{x} \\
\frac{2 x-1}{x-1}=2+\frac{1}{x-1}, & \frac{2 x+1}{x}=2+\frac{1}{x} .
\end{array}
$$

Therefore,

$$
K_{1}=\frac{-\frac{1}{x+1}+\frac{1}{x}}{\frac{1}{x-1}-\frac{1}{x}}=\frac{\frac{1}{x(x+1)}}{\frac{1}{(x-1) x}}=\frac{x-1}{x+1}=1-\frac{2}{x+1}
$$

(provided that $x \neq 0, \pm 1$, because in these cases the expression (1) is undefined).

II. The numerators and denominators in $K_{2}$ can easily be factored, and then the common factors in the terms can be simplified:

$$
\begin{gathered}
K_{2}=\frac{x(x+1)}{(x+1)(x+2)}-\frac{(x-1)(x-2)}{x(x-1)}+\frac{(x-3)(x-4)}{(x-2)(x-3)}= \\
=\frac{x}{x+2}-\frac{x-2}{x}+\frac{x-4}{x-2}=1-\frac{2}{x+2}-1+\frac{2}{x}+1-\frac{2}{x-2}= \\
=1-2\left(\frac{1}{x+2}-\frac{1}{x}+\frac{1}{x-2}\right) .
\end{gathered}
$$

The transformation was carried out in the manner of part I. If the value of the expression needs to be calculated for a small absolute value integer $x$ (naturally different from 0, $\pm 1, \pm 2$, and 3), it is practical to combine the middle term in the parentheses with either the first or the third term:

$$
K_{2}=1+\frac{4}{4(x+2)}-\frac{2}{x-2}=1-\frac{2}{x+2}-\frac{4}{x(x-2)}
$$

However, for large $x$, it is simpler to calculate using the last form above. Alternatively, everything can be brought to a common denominator:

$$
\begin{gathered}
K_{2}=\frac{x^{3}-2 x^{2}-4 x-8}{x\left(x^{2}-4\right)}=\frac{(x-2) x^{2}-4(x-2)-16}{x\left(x^{2}-4\right)}= \\
=\frac{(x-2)^{2}(x+2)-16}{x\left(x^{2}-4\right)}
\end{gathered}
$$

Balogh Irén (Gyula, Erkel F. g. II. o. t.)