Đáp án:

$\begin{array}{l}
B1)\\
1)Dkxd: - 2x - 6 \ge 0\\
 \Rightarrow 2x \le  - 6\\
 \Rightarrow x \le  - 3\\
2)a)\sqrt {{x^2} + 6x + 9}  = 7\\
 \Rightarrow \sqrt {{{\left( {x + 3} \right)}^2}}  = 7\\
 \Rightarrow \left| {x + 3} \right| = 7\\
 \Rightarrow \left[ \begin{array}{l}
x + 3 = 7\\
x + 3 =  - 7
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 10
\end{array} \right.\\
b)Dkxd:x \ge 1\\
\sqrt {25x - 25}  - \dfrac{{15}}{2}\sqrt {\dfrac{{x - 1}}{9}}  = 6 + \sqrt {x - 1} \\
 \Rightarrow \sqrt {25} .\sqrt {x - 1}  - \dfrac{{15}}{2}.\dfrac{{\sqrt {x - 1} }}{{\sqrt 9 }} = 6 + \sqrt {x - 1} \\
 \Rightarrow 5\sqrt {x - 1}  - \dfrac{{15}}{2}.\dfrac{1}{3}.\sqrt {x - 1}  - \sqrt {x - 1}  = 6\\
 \Rightarrow \left( {5 - \dfrac{5}{2} - 1} \right)\sqrt {x - 1}  = 6\\
 \Rightarrow \dfrac{3}{2}.\sqrt {x - 1}  = 6\\
 \Rightarrow \sqrt {x - 1}  = 4\\
 \Rightarrow x - 1 = 16\\
 \Rightarrow x = 17\left( {tmdk} \right)\\
BII)\\
a)2\sqrt 8  + \sqrt {32}  - \sqrt {200} \\
 = 2.2.\sqrt 2  + 4\sqrt 2  - 10\sqrt 2 \\
 =  - 2\sqrt 2 \\
b)\dfrac{1}{{\sqrt 5  + 2}} - \sqrt {9 + 4\sqrt 5 } \\
 = \dfrac{{\sqrt 5  - 2}}{{\left( {\sqrt 5  + 2} \right)\left( {\sqrt 5  - 2} \right)}} - \sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}} \\
 = \dfrac{{\sqrt 5  - 2}}{{5 - 4}} - \left( {\sqrt 5  + 2} \right)\\
 = \sqrt 5  - 2 - \sqrt 5  - 2\\
 =  - 4\\
c)2\sqrt[3]{8} - \sqrt[3]{{27}} + \sqrt[3]{{64}} + 3\sqrt[3]{{ - 125}}\\
 = 2.2 - 3 + 4 + 3.\left( { - 5} \right)\\
 = 4 - 3 + 4 - 15\\
 =  - 10\\
d)\dfrac{{\sin {{25}^0}}}{{\cos {{65}^0}}} + {\sin ^2}{15^0} - \left( {2021 - {{\cos }^2}{{15}^0}} \right)\\
 = \dfrac{{\sin {{25}^0}}}{{\sin {{25}^0}}} + {\sin ^2}{15^0} + {\cos ^2}{15^0} - 2021\\
 = 1 + 1 - 2021\\
 =  - 2019
\end{array}$