Question

Mn ơi, giúp e bài này với ạ, e cảm ơn trc nhé !? 
Tìm x:
3x.(x-2)-x²+2x=0

Answer 1

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

vậy x=0 và x=2

Answer 2

3x(x-2)-x^2+2x=0

<=>3x(x-2)-x(x-2)=0

<=>(3x-x)(x-2)=0

<=>2x(x-2)=0

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2

Answer 3

\(3x\left(x-2\right)-x^2+2x=0\Rightarrow3x\left(x-2\right)-x\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-x\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Answer 4

Ta có: \(3x\left(x-2\right)-x^2+2x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)