Tìm x biết \(\sqrt[3]{x-1}+1=x\)

* Hướng dẫn giải

Ta có

\(\begin{array}{l} \sqrt[3]{x-1}+1=x \\ \Leftrightarrow \sqrt[3]{x-1}=x-1 \Leftrightarrow(\sqrt[3]{x-1})^{3}=(x-1)^{3} \Leftrightarrow x-1=(x-1)^{3} \\ \Leftrightarrow(x-1)\left[(x-1)^{2}-1\right]=0 \Leftrightarrow\left[\begin{array}{c} x-1=0 \\ (x-1)^{2}-1=0 \end{array} \Leftrightarrow\left[\begin{array}{c} x=1 \\ (x-1)^{2}=1 \end{array}\right.\right. \end{array}\)

\(\Leftrightarrow \left[ \begin{array}{l} x = 1\\ \left[ \begin{array}{l} x - 1 = 1\\ x - 1 = - 1 \end{array} \right. \end{array} \right.\)

\(\Leftrightarrow\left[\begin{array}{l} x=1 \\ x=2 \\ x=0 \end{array}\right.\)

Vậy tập nghiệm \(S=\{0 ; 1 ; 2\}\)