Question

решите......................

Answer 1

✔ответ ✔

●: 1) $\frac{3+xy}{3y} - \frac{x-3y}{xy} = \frac{x*(3+xy)-3(x-3y)}{3xy} = \frac{3x+x^2y-3x+9y}{3xy} = \frac{y*(x^2+9)}{3xy} = \frac{x^2+9}{3x}$

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

3) $\frac{1-5a^2}{a^6} - \frac{a-5}{a^4} - \frac{1}{a^3} = \frac{1-5a^2-a^2*(a-5)-a^3}{a^6} = \frac{1-5a^2-a^3+5a^2-a^3}{a^6} = \frac{1-2a^3}{a^6}$

5) $\frac{x}{x^2-3x} + \frac{x}{9-3x} = \frac{x}{x*(x-3)} + \frac{x}{-3x+9} = \frac{1}{x-3} + \frac{x}{-3(x-3)} = \frac{1}{x-3} - \frac{x}{3(x-3)} = \frac{3-x}{3(x-3)} = \frac{-(x-3)}{3(x-3)} = \frac{-1}{3} = -\frac{1}{3} = - 0,3$