Giải thích các bước giải:
DKXD : $y\ne \pm 5$
Ta có :
$\dfrac{y+1}{y-5}+\dfrac{10}{y+5}=\dfrac{y+1}{y-5}.\dfrac{10}{y+5}$
$\to \dfrac{(y+5)(y+1)+10(y-5)}{(y-5)(y+5)}=\dfrac{10(y+1)}{(y-5)(y+5)}$
$\to \dfrac{y^2+16y-45}{(y-5)(y+5)}=\dfrac{10(y+1)}{(y-5)(y+5)}$
$\to y^2+16y-45=10y+10$
$\to y^2+6y-55=0$
$\to y^2-5y+11y-55=0$
$\to (y-5)(y+11)=0$
$\to y=-11$ vì $y\ne 5$