a/=> x2 - 3x + x - 3 = 0
=> x(x-3) + (x-3) = 0
=> (x+1)(x-3) = 0
=> x + 1 = 0 => x = -1
hoặc x - 3 = 0 => x = 3
b/ => 2x2 - x + 6x - 3 = 0
=> x (2x-1) + 3 (2x-1) = 0
=> (x+3) (2x-1) = 0
=> x + 3 = 0 => x = -3
hoặc 2x - 1 = 0 => 2x = 1 => x = 1/2
a) x^2 - 2x - 3 = 0
<=>x2+x-3x-3=0
<=>x.(x+1)-3.(x+1)=0
<=>(x+1)(x-3)=0
<=>x+1=0 hoặc x-3=0
<=>x=-1 hoặc x=3
b) 2x^2 + 5x - 3 = 0
<=>2x2-x+6x-3=0
<=>x.(2x-1)+3.(2x-1)=0
<=>(2x-1)(x+3)=0
<=>2x-1=0 hoặc x+3=0
<=>x=1/2 hoặc x=-3