Question

решите пожалуйста алгебра 8класс дам 25 баллов

Answer 1

$\dfrac{4y+24}{5y^2-45}+\dfrac{y+3}{5y^2-15y}=\dfrac{y-3}{y^2+3y}$

$\dfrac{4y+24}{5(y^2-9)}+\dfrac{y+3}{5y(y-3)}-\dfrac{y-3}{y(y+3)}=0$

$\dfrac{4y+24}{5(y-3)(y+3)}+\dfrac{(y+3)^2-5(y-3)^2}{5y(y-3)(y+3)}=0$

$\dfrac{4y+24}{5(y-3)(y+3)}+\dfrac{y^2+6y+9-5y^2+30y-45}{5y(y-3)(y+3)}=0$

$\dfrac{(4y+24)y-4y^2+36y-36}{5y(y-3)(y+3)}=0$

$\dfrac{4y^2+24y-4y^2+36y-36}{5y(y-3)(y+3)}=0$

$\dfrac{60y-36}{5y(y-3)(y+3)}=0;\quad ODZ:y\neq0;\ 3;\ -3$

$60y-36=0\\\\60y=36\\\\y=\dfrac{36}{60}\\\\y=0.6$

Ответ:  y = 0,6