Question

Упростите выражение
$frac{1,5x}{x+y} + frac{x-y}{x^2+2xy}:( frac{y}{x^2-2xy}- frac{2x+y}{x^2-4y^2}$

Answer 1

$frac{1,5x}{x+y} + frac{x-y}{x^2+2xy}:( frac{y}{x^2-2xy}- frac{2x+y}{x^2-4y^2} )= \ =frac{1,5x}{x+y} + frac{x-y}{x(x+2y)}:( frac{y}{x(x-2y)}- frac{2x+y}{(x-2y)(x+2y)} )= \ =frac{1,5x}{x+y} + frac{x-y}{x(x+2y)}:frac{y(x+2y)-(2x+y)x}{x(x-2y)(x+2y)}= \ =frac{1,5x}{x+y} + frac{x-y}{x(x+2y)}:frac{xy+2y^2-2x^2-xy}{x(x-2y)(x+2y)}= \ =frac{1,5x}{x+y} + frac{x-y}{x(x+2y)}:frac{2(y-x)(y+x)}{x(x-2y)(x+2y)}=$
$=frac{1,5x}{x+y} + frac{x-y}{x(x+2y)}cdotfrac{x(x-2y)(x+2y)}{2(y-x)(y+x)}= \ =frac{1,5x}{x+y} -frac{(x-y)x(x-2y)(x+2y)}{2(x-y)(y+x)x(x+2y)}= \ =frac{3x}{2(x+y)} -frac{x-2y}{2(y+x)}=frac{3x-x+2y}{2(x+y)}=frac{2x+2y}{2x+2y}=1$