a)5x(x−2000)−x+2000=05x(x−2000)−(x−2000)=0(x−2000)(5x−1)=0⇒[x−2000=05x−1=0⇒[x=2000x=15a)5x(x−2000)−x+2000=05x(x−2000)−(x−2000)=0(x−2000)(5x−1)=0⇒[x−2000=05x−1=0⇒[x=2000x=15
Vậy x=15x=15 hoặc x=2000x=2000
b)x3−13x=0x(x2−13)=0⇒[x=0x2−13=0⇒[x=0x=±√13b)x3−13x=0x(x2−13)=0⇒[x=0x2−13=0⇒[x=0x=±13
Vậy x=0x=0 hoặc x=±√13