7. For which integer values of $k$ does the quadratic equation $k x^{2}+(2 k-1) x+k-2=0$ have (20) rational solutions.
Solution. It is necessary and sufficient that the discriminant

$$
D=(2 k-1)^{2}-4 k(k-2)=4 k+1
$$

be the square of a rational number $m$, i.e., $4 k+1=m^{2}$. From this, we see that $m$ must be an odd integer.

(5 points)

Since $m$ is odd, i.e., $m=2 a+1$, we have

$$
\begin{aligned}
4 k+1 & =(2 a+1)^{2} \\
4 k+1 & =4 a^{2}+4 a+1 \\
k & =a^{2}+a=a(a+1)
\end{aligned}
$$

(15 points)

Therefore, the number $k$ must be of the form $a^{2}+a, a \in \mathbb{Z}$, or $a(a+1)$, i.e., equal to the product of two consecutive integers. The discriminant is $D=4 k+1=(2 a+1)^{2}$.