Question

\frac{2x +y}{2 x^{2} -xy} + \frac{16x}{4 x^{2} - y^{2} } - \frac{2x-y}{2 x^{2} + xy }

Answer 1

$\frac{2x +y}{2 x^{2} -xy} + \frac{16x}{4 x^{2} - y^{2} } - \frac{2x-y}{2 x^{2} + xy }= \frac{2x+y}{x(2x-y)} + \frac{16x}{(2x-y)(2x+y)}- \frac{2x-y}{x(2x+y)}= \\ = \frac{(2x+y)^2+16x^2-(2x-y)^2}{x(x-2y)(x+2y)}= \frac{4x^2+4xy+y^2+16x^2-4x^2+4xy-y^2}{x(2x-y)(2x+y)}= \frac{16x^2+8xy}{x(2x-y)(2x+y)} =\\ = \frac{8x(2x+y)}{x(2x-y)(2x+y)}= \frac{8}{2x-y}$