Answer
$4x-\displaystyle \frac{2}{y}-\frac{5}{xy^{2}}$
Work Step by Step
Using the property: $\displaystyle \quad\frac{A+B}{C}=\frac{A}{C}+\frac{B}{C},$
$\displaystyle \frac{40x^{4}y^{3}-20x^{3}y^{2}-50x^{2}y}{10x^{3}y^{3}}=\frac{40x^{4}y^{3}}{10x^{3}y^{3}}-\frac{20x^{3}y^{2}}{10x^{3}y^{3}}-\frac{50x^{2}y}{10x^{3}y^{3}}$
= $4x-\displaystyle \frac{2}{y}-\frac{5}{xy^{2}}$
