Question

Giải PT bằng cách đưa về PT tích

\(\dfrac{\left(x-3\right)^3}{2}=x^2-6x+9\)

Answer 1

\(\Leftrightarrow\dfrac{\left(x-3\right)^3}{2}=\left(x-3\right)^2\)

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

\(\Leftrightarrow\left(x-3\right)^2\left[\left(x-3\right)-2\right]=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\x-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)