Question

giúp mình với ạ
Tìm x:

a)x(x-3)-2x+6=0
b) 5(x-1)-x^2+x=0

Answer 1

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 \(x=1\)

Answer 2

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