Đáp án:
b. x=1
Giải thích các bước giải:
\(\begin{array}{l}
a.\sqrt[3]{{2x + 1}} + \sqrt[3]{x} = 1\\
\to \sqrt[3]{{2x + 1}} = 1 - \sqrt[3]{x}\\
\to 2x + 1 = 1 - 3.1.\sqrt[3]{x} + 3.\sqrt[3]{{{x^2}}} - x\\
\to 3x = - 3\sqrt[3]{x} + 3.\sqrt[3]{{{x^2}}}\\
\to x = \sqrt[3]{{{x^2}}} - \sqrt[3]{x}\\
\to \sqrt[3]{{{x^3}}} = \sqrt[3]{{{x^2}}} - \sqrt[3]{x}\\
\to \sqrt[3]{{{x^3}}} - \sqrt[3]{{{x^2}}} + \sqrt[3]{x} = 0\\
\to \sqrt[3]{x}\left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1} \right) = 0\\
\to x = 0\\
b.DK:x \ge \dfrac{1}{2}\\
\sqrt {x + \sqrt {2x - 1} } + \sqrt {x - \sqrt {2x - 1} } = \sqrt 2 \\
\to \sqrt {2x - 1 + 2\sqrt {2x - 1} .1 + 1} + \sqrt {2x - 1 - 2\sqrt {2x - 1} .1 + 1} = 2\\
\to \sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} = 2\\
\to \left| {\sqrt {2x - 1} + 1} \right| + \left| {\sqrt {2x - 1} - 1} \right| = 2\\
\to \left[ \begin{array}{l}
\sqrt {2x - 1} + 1 + \sqrt {2x - 1} - 1 = 2\\
\sqrt {2x - 1} + 1 - \sqrt {2x - 1} + 1 = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt {2x - 1} = 2\\
2 = 2\left( {ld} \right)
\end{array} \right.\\
\to \sqrt {2x - 1} = 1\\
\to 2x - 1 = 1\\
\to x = 1\left( {TM} \right)
\end{array}\)
