a.(x-2)*5x-(2-x)=0
=>(x-2)*5x+(x-2)=0
=> (x-2)*(5x+1)=0
Th1: x-2 =0 suy ra x=2
Th2: 5x+1 =0 suy ra x= \(-\frac{1}{5}\)
b. 4x(x+1)= 8(x+1)
=>(4x-8)(x-1)=0
Th1:4x-8 =0 suy ra x= 2
Th2: x-1=0 suy ra x=1
c.x^3=x^5
=>x^3-x^5=0
=>x^3(1-x^2)=0
Th1:x^3=0 suy ra x=0
Th2:1-x^2=0 suy ra x=1; x=-1.