Question

Алгеброические дроби, хелп
(c права буду писать степени)(/ - дробная черта)( x - умножить)
y2 - 25/y + 3 x 1/y2 + 5y - y+5/y2 - 3y


буду признателен

Answer 1

1)(y-5)(y+5)/(y+3)*1/[y(y+5)]=(y-5)/[y(y+3)]
2)(y-5)/[y(y+3)-(y+5)/[y(y-3)]=[(y-3)(y-5)-(y+3)(y+5)]/[y(y²-9)]=
=(y²-3y-5y+15-y²-3y-5y-15)/[y(y²-9)]=-16y/[y(y²-9)]=16/(9-y²)