Đáp án:

\(\begin{array}{l}
B1:\\
\sqrt {2019}  + \sqrt {2021}  > 2.\sqrt {2020} \\
B2:\\
a)\sqrt 6  - 22\\
b)33\\
c)2\\
d)4\sqrt 3 \\
B3:\\
a)x = 32\\
b)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 4\\
x = 9
\end{array} \right.\\
d)\left[ \begin{array}{l}
x =  - 3\\
x =  - 2,971960768\\
x = 3,395426035\\
x = 2,576534734
\end{array} \right.
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
{\left( {\sqrt {2019}  + \sqrt {2021} } \right)^2} = 2019 + 2\sqrt {2019.2021}  + 2021\\
 = 4040 + 2\sqrt {2019.2021} \\
{\left( {2.\sqrt {2020} } \right)^2} = 2.2020 = 4040\\
Do:4040 + 2\sqrt {2019.2021}  > 4040\\
 \to {\left( {\sqrt {2019}  + \sqrt {2021} } \right)^2} > {\left( {2.\sqrt {2020} } \right)^2}\\
 \to \sqrt {2019}  + \sqrt {2021}  > 2.\sqrt {2020} \\
B2:\\
a)\dfrac{3}{4}.\sqrt 2 .\left( {\dfrac{4}{3}.\sqrt 3  - \dfrac{2}{3}.4\sqrt 2  - 4.3\sqrt 2 } \right)\\
 = \dfrac{3}{{2\sqrt 2 }}.\left( {\dfrac{4}{{\sqrt 3 }} - \dfrac{{8\sqrt 2 }}{3} - 12\sqrt 2 } \right)\\
 = \sqrt 6  - 4 - 18 = \sqrt 6  - 22\\
b)\left( {7.4\sqrt 3  + 3.3\sqrt 3  - 2.2\sqrt 3 } \right):\sqrt 3 \\
 = \dfrac{{\left( {28 + 9 - 4} \right)\sqrt 3 }}{{\sqrt 3 }} = 33\\
c)\sqrt {5 + 2\sqrt 5 .1 + 1}  - \sqrt {5 - 2\sqrt 5 .1 + 1} \\
 = \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}} \\
 = \sqrt 5  + 1 - \sqrt 5  + 1 = 2\\
d)\sqrt {12 + 2.2\sqrt 3 .3 + 9}  + \sqrt {12 - 2.2\sqrt 3 .3 + 9} \\
 = \sqrt {{{\left( {2\sqrt 3  + 3} \right)}^2}}  + \sqrt {{{\left( {2\sqrt 3  - 3} \right)}^2}} \\
 = 2\sqrt 3  + 3 + 2\sqrt 3  - 3 = 4\sqrt 3 \\
B3:\\
a)DK:x \ge 0\\
\sqrt {2x}  = 8\\
 \to 2x = 64\\
 \to x = 32\\
b){\sqrt {\left( {2x - 1} \right)} ^2} = 2\\
 \to \left| {2x - 1} \right| = 2\\
 \to \left[ \begin{array}{l}
2x - 1 = 2\\
2x - 1 =  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
c)DK:x \ge 0\\
x - 5\sqrt x  + 6 = 0\\
 \to \left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right) = 0\\
 \to \left[ \begin{array}{l}
\sqrt x  - 2 = 0\\
\sqrt x  - 3 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 4\\
x = 9
\end{array} \right.\\
d)DK:x \ge  - 3\\
\left( {x - 3} \right)\left( {x + 3} \right) - \sqrt {x + 3}  = 0\\
 \to \sqrt {x + 3} \left( {\sqrt {x + 3} \left( {x - 3} \right) - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x + 3 = 0\\
\sqrt {x + 3} \left( {x - 3} \right) - 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\\
\sqrt {x + 3} \left( {x - 3} \right) = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\\
\left( {x + 3} \right)\left( {{x^2} - 6x + 9} \right) = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\\
{x^3} - 6{x^2} + 9x + 3{x^2} - 18x + 27 - 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\\
x =  - 2,971960768\\
x = 3,395426035\\
x = 2,576534734
\end{array} \right.
\end{array}\)