HELP .-. Giải phương trình sau : 120/x = 120/x+10 + 0,6

ĐKXĐ: `x ne 0; x ne -10`
    `120/x=120/(x+10)+0,6`

`<=>120/x=120/(x+10)+6/10`

`<=>[120.10.(x+10)]/[10x.(x+10)]=(120.10x)/[10x.(x+10)]+[6x.(x+10)]/[10x.(x+10)]`

`<=>1200.(x+10)=1200x+6x.(x+10)`

`<=>1200x+12000=1200x+6x^2+60x`

`<=>1200x-1200x=6x^2+60x-12000`

`<=>6x^2+60x-12000=0`

`<=>6.(x^2+10x-2000)=0`

`<=>(x^2+10x-2000)=0`

`<=>x^2+50x-40x-2000=0`

`<=>x.(x+50)-40.(x+50)=0`

`<=>(x+50).(x-40)=0`

`<=>`$\left[\begin{matrix} x+50=0\\ x-40=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-50 (nhận)\\ x=40 (nhận)\end{matrix}\right.$

Vậy `S={-50;40}`