Lần sau em tách hỏi từng câu thôi nhé
\(\begin{array}{l}
D = \frac{1}{{\sqrt 2 }}\left( {\sqrt {10 + 2\sqrt {21} } + \sqrt {10 - 2\sqrt {21} } } \right)\\
= \frac{1}{{\sqrt 2 }}\left( {\sqrt {7 + 2\sqrt 7 \sqrt 3 + 3} - \sqrt {7 - 2\sqrt 7 \sqrt 3 + 3} } \right)\\
= \frac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} } \right)\\
= \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 + \sqrt 3 - \sqrt 7 + \sqrt 3 } \right)\\
= \frac{1}{{\sqrt 2 }}.2\sqrt 3 = \sqrt 6 \\
6)\, = \sqrt {xy} + \sqrt x + \sqrt y + 1\\
= \sqrt x \left( {\sqrt y + 1} \right) + \left( {\sqrt y + 1} \right)\\
= \left( {\sqrt y + 1} \right)\left( {\sqrt x + 1} \right)\\
7)\,\sqrt x = x\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow {x^2} = x\\
\Leftrightarrow x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\left( {tm} \right)\\
\sqrt[3]{{2x - 4}} = 2\\
2x - 4 = 8\\
2x = 12\\
x = 6\\
8)\,A = 4 - \sqrt {{{\left( {x - 2} \right)}^2}} \\
= 4 - \left| {x - 2} \right| \le 4\\
{A_{\max }} = 4 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2
\end{array}\)
