Question

giải phương trình sau:
x(x-4)-3x+12=0

 

Answer 1

Answer 2

`x(x - 4) - 3x + 12 = 0`

`<=> x(x - 4) + 3(x - 4) = 0`

`<=> (x + 3)(x - 4) = 0`

`<=>` $\left[\begin{matrix} x + 3 = 0\\ x - 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$

Vậy `S = {-3; 4}`

Answer 3

\(x\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)