(x - 2)2 - 3x + 6 = 0
=> (x - 2)2 - 3.(x - 2) = 0
=> (x - 2).(x - 2 - 3) = 0
=> (x - 2).(x - 5) = 0
=> \(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
Vậy \(x\in\left\{2;5\right\}\)
(x - 2)2 - 3x + 6 = 0
=> (x - 2)2 - 3(x - 2) = 0
=> (x - 2).(x - 2 - 3) = 0
=> (x - 2)(x - 5) = 0
=> x - 2 = 0 => x = 2
hoặc x - 5 = 0 => x = 5
Vậy x = 2, x = 5
(x - 2)2 - 3x + 6 = 0
=> (x - 2)2 - 3.(x - 2) = 0
=> (x - 2).(x - 2 - 3) = 0
=> (x - 2).(x - 5) = 0
=> \(\orbr{\begin{cases}x-2\\x-5\end{cases}=0}\)
| x=2 |
| x=5 |
Vậy x∈{2;5}
(x - 2)2 - 3x + 6 = (x - 2)2 - (3x - 6) = (x - 2)2 - 3(x - 2) = (x - 2)(x - 2 - 3) = (x - 2)(x - 5) = 0
=> x - 2 = 0 hoặc x - 5 = 0 => x = 2 ; 5
x - 2)2 - 3x + 6 = 0
=> (x - 2)2 - 3.(x - 2) = 0
=> (x - 2).(x - 2 - 3) = 0
=> (x - 2).(x - 5) = 0
=> $$
Vậy $x\in\left\{2;5\right\}$
