Question

Доведіть тотожність:

Answer 1

Ответ:

$\displaystyle \left(\frac{20y}{y^2-25}+\frac{2y}{y+5}-\frac{2y}{15-3y}\right):\frac{2y}{3y-15}=4$

Объяснение:

$\displaystyle \left(\frac{20y}{y^2-25}+\frac{2y}{y+5}-\frac{2y}{15-3y}\right):\frac{2y}{3y-15}=\\\\\\\left(\frac{20y}{(y-5)(y+5)}+\frac{2y}{y+5}-\frac{2y}{-3(y-5)}\right):\frac{2y}{3(y-5)}=\\\\\\\left(\frac{20y}{(y-5)(y+5)}+\frac{2y}{y+5}+\frac{2y}{3(y-5)}\right)\cdot \frac{3(y-5)}{2y}=\\\\\\\left(\frac{20y\cdot 3}{3(y-5)(y+5)}+\frac{2y\cdot 3(y-5)}{3(y-5)(y+5)}+\frac{2y\cdot(y+5)}{3(y-5)(y+5)}\right)\cdot \frac{3(y-5)}{2y}=$

$\displaystyle \frac{60y+6y(y-5)+2y(y+5)}{3(y-5)(y+5)}\cdot \frac{3(y-5)}{2y}=\\\\\\\frac{60y+6y^2-30y+2y^2+10y}{3(y-5)(y+5)}\cdot \frac{3(y-5)}{2y}=\\\\\\\frac{8y^2 + 40y}{y+5}\cdot \frac{1}{2y}=\\\\\\\frac{8y(y+5)}{y+5}\cdot \frac{1}{2y}=4$