Question

Правильно ли я решил? ! 5/b+2 - 1/b = 5b-b+2/b²+2 = 4b+2/b²+2 x²-3xy/x²-y² + y/x-y = x²-3xy=y(x-y)/x²-y² = x²-3xy+xy-y²/x²-y² = x²-2xy-y²/x²-y² /-дробная черта

Answer 1

$\frac{5}{b+2}- \frac{1}{b}= \frac{5*b-1*(b+2)}{(b+2)*b}= \frac{5b-b-2}{(b+2)*b}= \frac{4b-2}{b(b+2)} \\ \frac{x^2-3xy}{x^2-y^2}+ \frac{y}{x-y}=\frac{x^2-3xy}{(x-y)(x+y)}+ \frac{y}{x-y}=\frac{x^2-3xy+y(x+y)}{(x-y)(x+y)}=\frac{x^2-3xy+xy+y^2}{(x-y)(x+y)}= \\ =\frac{x^2-2xy+y^2}{(x-y)(x+y)}=\frac{(x-y)^2}{(x-y)(x+y)}= \frac{x-y}{x+y}$