Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 1\\
M = \dfrac{{x + \sqrt[3]{x} - 2}}{{x - 1}} - \dfrac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}} + \dfrac{1}{{\sqrt[3]{x} - 1}}\\
= \dfrac{{x + \sqrt[3]{x} - 2}}{{\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}} - \dfrac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}}\\
+ \dfrac{1}{{\sqrt[3]{x} - 1}}\\
= \dfrac{{x + \sqrt[3]{x} - 2 - \sqrt[3]{x} + 1 + \sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}}{{\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
= \dfrac{{x + \sqrt[3]{{{x^2}}} + \sqrt[3]{x}}}{{\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
= \dfrac{{\sqrt[3]{x}\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}{{\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
= \dfrac{{\sqrt[3]{x}}}{{\sqrt[3]{x} - 1}}\\
b)M = \dfrac{1}{3}\\
\Rightarrow \dfrac{{\sqrt[3]{x}}}{{\sqrt[3]{x} - 1}} = \dfrac{1}{3}\\
\Rightarrow 3\sqrt[3]{x} = \sqrt[3]{x} - 1\\
\Rightarrow \sqrt[3]{x} = \dfrac{{ - 1}}{2}\\
\Rightarrow x = {\left( { - \dfrac{1}{2}} \right)^3}\\
\Rightarrow x = \dfrac{{ - 1}}{8}\left( {tmdk} \right)
\end{array}$
