Question

giải phương trình \(\sqrt[3]{x-1}+1=x\)

Answer 1

\(Đkx\ge1\)

Đặt \(\sqrt[3]{x-1}=a\)

\(\Rightarrow a=a^3\)

\(\Leftrightarrow a\left(a^2-1\right)=0\Leftrightarrow a\left(a-1\right)\left(a+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x-1}=0\\\sqrt[3]{x-1}=1\\\sqrt[3]{x-1}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhạn\right)\\x=2\left(nhận\right)\\x=0\left(loại\right)\end{matrix}\right.\)

Vậy S=\(\left\{1,2\right\}\)