a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x-3=0\)
\(x=2\) hoặc \(x=3\)
b) \(5\left(x-1\right)-x^2+x=0\)
\(5\left(x-1\right)-x\left(x-1\right)=0\)
\(\left(5-x\right)\left(x-1\right)=0\)
\(\Rightarrow5-x=0\) hoặc \(x-1=0\)
\(x=5\) hoặc \(x=1\)
a) x(x - 3) - 2x + 6 = 0
x(x - 3) - (2x - 6) = 0
x(x - 3) - 2(x - 3) = 0
(x - 3)(x - 2) = 0
x - 3 = 0 hoặc x - 2 = 0
*) x - 3 = 0
x = 3
*) x - 2 = 0
x = 2
Vậy x = 2; x = 3
b) 5(x - 1) - x² + x = 0
5(x - 1) - (x² - x) = 0
5(x - 1) - x(x - 1) = 0
(x - 1)(5 - x) = 0
x - 1 = 0 hoặc 5 - x = 0
*) x - 1 = 0
x = 1
*) 5 - x = 0
x = 5
Vậy x = 1; x = 5
