a,8(1-x)³=27
⇔(1-x)³=$\frac{27}{8}$
⇔1-x=$\sqrt[3]{\frac{27}{8}}$
⇔x=$\frac{-1}{2}$
b,x=3$\sqrt[]{x}$ (Đk x≥0)
⇔x-3$\sqrt[]{x}$ =0
⇔$\sqrt[]{x}$ .($\sqrt[]{x}$ -3)=0
⇔\(\left[ \begin{array}{l}\sqrt[]{x}=0\\\sqrt[]{x}-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=9\end{array} \right.\)
