Giải thích các bước giải:
a,
ĐKXĐ: \(x \ge - 3\)
Ta có:
\(\begin{array}{l}
\sqrt {x + 3} - 2\sqrt {2x + 7} + 4 = 0\\
\Leftrightarrow \left( {\sqrt {x + 3} - 2} \right) + \left( {6 - 2\sqrt {2x + 7} } \right) = 0\\
\Leftrightarrow \frac{{x + 3 - 4}}{{\sqrt {x + 3} + 2}} + 2.\frac{{9 - \left( {2x + 7} \right)}}{{3 + \sqrt {2x + 7} }} = 0\\
\Leftrightarrow \frac{{x - 1}}{{\sqrt {x + 3} + 2}} + \frac{{4\left( {1 - x} \right)}}{{3 + \sqrt {2x + 7} }} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {\frac{1}{{\sqrt {x + 3} + 2}} - \frac{4}{{3 + \sqrt {2x + 7} }}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
4\sqrt {x + 3} + 8 = 3 + \sqrt {2x + 7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\sqrt {16x + 48} + 5 - \sqrt {2x + 7} =
\end{array} \right.\\
x \ge - 3 \Rightarrow \sqrt {16x + 48} + 5 > \sqrt {2x + 7} \\
\Rightarrow x = 1
\end{array}\)
b,
\(\begin{array}{l}
3\sqrt {x - 1} + \sqrt[3]{{3x + 2}} = 5\\
\Leftrightarrow 3\left( {\sqrt {x - 1} - 1} \right) + \left( {\sqrt[3]{{3x + 2}} - 2} \right) = 0\\
\Leftrightarrow 3.\frac{{x - 2}}{{\sqrt {x - 1} + 1}} + \frac{{3x - 6}}{{\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 2\sqrt[3]{{3x + 2}} + 4}} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
\frac{1}{{\sqrt {x - 1} + 1}} + \frac{1}{{\sqrt[3]{{{{\left( {3x + 2} \right)}^2}}} + 2\sqrt[3]{{3x + 2}} + 4}} = 0\,\,\,\,\,\left( {VN} \right)
\end{array} \right.
\end{array}\)
Vậy \(x = 2\)
