Question

Решите, умоляю х3+27у3 9у2+6ху+х2 20

Answer 1

$\frac{ {x}^{3} + 27 {y}^{3} }{9 {y}^{2} + 6xy + {x}^{2} } = \\ = \frac{ {x}^{3} + (3y) ^{3} }{ {(3y)}^{2} + 2 \cdot x\cdot(3y) + {x}^{2} } = \\ = \frac{(x + 3y)( {x}^{2} - x\cdot (3y) + 9 {y}^{2}) }{(3y + x) ^{2} } = \\ = \frac{{x}^{2} - 3xy + 9 {y}^{2}}{x + 3y}$