Question

Es otra Divicion Algebraica

Answer 1

$\frac{\frac{1}{4}x^5y^2+\frac{2}{3}x^2y^3-\frac{3}{4}x^4y^3}{\frac{1}{2}xy^2}\\= \frac{\frac{1}{4}x^5y^2}{\frac{1}{2}xy^2}+\frac{\frac{2}{3}x^2y^3}{\frac{1}{2}xy^2}-\frac{\frac{3}{4}x^4y^3}{\frac{1}{2}xy^2} \\=\frac{2}{4}x^4+\frac{4}{3}xy-\frac{6}{4}x^3y\\ =\frac{1}{2}x^4+\frac{4}{3}xy-\frac{3}{2}x^3y$