The number of values of k for which $\left[{\mathrm{x}}^2-(\mathrm{k}-2)\mathrm{x}+{\mathrm{k}}^2\right]\times \left[{\mathrm{x}}^2+\mathrm{kx}+(2\mathrm{k}-1)\right]$ is a perfect square is

For given situation, ${\mathrm{x}}^2-(\mathrm{k}-2)\mathrm{x}+{\mathrm{k}}^2=0\text{ and }{\mathrm{x}}^2+\mathrm{kx}+2\mathrm{k}-1=0$

should have both roots common or each should have equal
roots. If both roots are common, then

$\frac{1}{1}=\frac{-(\mathrm{k}-2)}{\mathrm{k}}=\frac{{\mathrm{k}}^2}{2\mathrm{k}-1}$

$\Rightarrow \mathrm{k}=-\mathrm{k}+2\text{ and }2\mathrm{k}-1={\mathrm{k}}^2\Rightarrow \mathrm{k}=1$

If both the equations have equal roots, then

$\begin{array}{l} (\mathrm{k}-2{)}^2-4{\mathrm{k}}^2=0\text{ and }{\mathrm{k}}^2-4(2\mathrm{k}-1)=0\\ \Rightarrow (3\mathrm{k}-2)(-\mathrm{k}-2)=0\text{ and }{\mathrm{k}}^2-8\mathrm{k}+4=0\end{array}$

There is no common value of k.
Therefore, k = 1 is the only possible value.