If $\frac{2x-1}{\left(x-1\right)\left(2x+3\right)}=\frac{1}{5\left(x-1\right)}-\frac{k}{5\left(2x+3\right)}$  then $\left|k\right|=$

$\frac{2x-1}{\left(x-1\right)\left(2x+3\right)}=\frac{2x+3-k\left(x-1\right)}{5\left(x-1\right)\left(2x+3\right)}$

$5\left(2x-1\right)=2x+3-kx+k$

$10x-5=\left(2-k\right)x+3+k$

$2-k=10\text{or }3+k=-5$

$2-10=k\text{or }k=-5-3$

$k=-8\text{or }k=-8$

$\left|k\right|=8or\left|k\right|=8$