Question

[Nguyễn Đức Lâm]

help me căn bậc ba x-2 = x-2 

  tìm x

Answer 1

[Nguyễn Hoàng Minh]

\(\Leftrightarrow x-2=\left(x-2\right)^3\\ \Leftrightarrow\left(x-2\right)\left(x^2-4x+4-1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-4x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Answer 2

[Lấp La Lấp Lánh]

\(\sqrt[3]{x-2}=x-2\)

\(\Leftrightarrow\sqrt[3]{x-2}\left(\sqrt[3]{\left(x-2\right)^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)