Question

Tìm x biết :

x(x-2)² + 6(2-x)³ = 0

Answer 1

x.(x-2)\(^2\)-6.(x-2)\(^3\)=0

(x-2)\(^2\).\(\left[x-6.\left(x-2\right)\right]\)=0

(x-2)\(^2\).(x-6x+12)=0

(x-2)\(^2\).(-5x+12)=0

\(\left[{}\begin{matrix}\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\\-5x+12=0\Rightarrow-5x=-12\Rightarrow x=\frac{12}{5}\end{matrix}\right.\)

Answer 2

ta có \(x\left(x-2\right)^2+6\left(2-x\right)^3=0\)

⇔\(x\left(x-2\right)^2-6\left(x-2\right)^3=0\)

⇔\(\left(x-2\right)^2\left[x-6\left(x-2\right)\right]=0\)

⇔\(\left(x-2\right)^2\left[x-6x+2\right]=0\)

⇔\(\left(x-2\right)^2\left(-5x+2\right)=0\)

⇔\(\left[{}\begin{matrix}\left(x-2\right)^2=0\\-5x+2=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x-2=0\\-5x=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\5x=2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\x=\frac{2}{5}\end{matrix}\right.\)

vậy \(x\in\left\{2;\frac{2}{5}\right\}\)

Answer 3

Ta có:

\(x\left(x-2\right)^2+6\left(2-x\right)^3=0\)

\(x\left(x-2\right)^2-6\left(x-2\right)^3=0\)

\(\left(x-2\right)^2\left[x-6\left(x-2\right)\right]=0\)

\(\left(x-2\right)^2\left[x-6x+2\right]=0\)

\(\left(x-2\right)^2\left(-5x+2\right)=0\)

\(\left[{}\begin{matrix}\left(x-2\right)^2=0\\-5x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\5x=2\end{matrix}\right.\Leftrightarrow}}\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\)

Vậy:\(x\in\left\{2;\dfrac{2}{5}\right\}\)