Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
DK:x \ge - 1\\
\sqrt[3]{{x - 2}} + \sqrt {x + 1} = 3\\
\Leftrightarrow \left( {\sqrt[3]{{x - 2}} - 1} \right) + \left( {\sqrt {x + 1} - 2} \right) = 0\\
\Leftrightarrow \frac{{\left( {\sqrt[3]{{x - 2}} - 1} \right)\left( {\sqrt[3]{{{{\left( {x - 2} \right)}^2}}} + \sqrt[3]{{x - 2}} + 1} \right)}}{{\left( {\sqrt[3]{{{{\left( {x - 2} \right)}^2}}} + \sqrt[3]{{x - 2}} + 1} \right)}} + \frac{{\left( {\sqrt {x + 1} - 2} \right)\left( {\sqrt {x + 1} + 2} \right)}}{{\sqrt {x + 1} + 2}} = 0\\
\Leftrightarrow \frac{{x - 2 - 1}}{{\left( {\sqrt[3]{{{{\left( {x - 2} \right)}^2}}} + \sqrt[3]{{x - 2}} + 1} \right)}} + \frac{{x + 1 - 4}}{{\sqrt {x + 1} + 2}} = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {\frac{1}{{\left( {\sqrt[3]{{{{\left( {x - 2} \right)}^2}}} + \sqrt[3]{{x - 2}} + 1} \right)}} + \frac{1}{{\sqrt {x + 1} + 2}}} \right) = 0\\
\Leftrightarrow x = 3
\end{array}\]
