sửa lại:
a) \(5x\left(x-1\right)=x-1\)
=> \(5x\left(x-1\right)-\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(5x-1\right)=0\)
=> x - 1 = 0 hoặc 5x - 1 = 0
=> x = 1 hoặc 5x = 1 => x = 1/5
Vậy x = 1 hoặc x = 1/5
b) \(2\left(5+x\right)-x^2-5x=0\)
=> \(2\left(5+x\right)-\left(x^2+5x\right)=0\)
=> \(2\left(5+x\right)-x\left(x+5\right)=0\)
=> \(\left(x+5\right)\left(2-x\right)=0\)
=> x + 5 = 0 hoặc 2 - x = 0
=> x = -5 hoặc x = 2
a, 5x(x-1)=x-1
<=>5x(x-1)-(x-1)=0
<=>(5x-1)(x-1)=0
<=>5x-1=0 hoặc x-1=0
<=>x=1/5 hoặc x=1
b,2(5+x)-x2-5x=0
<=>2(x+5)-x(x+5)=0
<=>(2-x)(x+5)=0
<=>2-x=0 hoặc x+5=0
<=>x=2 hoặc x=-5
a, 5x(x-1)=x-1
5x = (x-1):(x-1)=1
=> x=1/5
:)
