Đáp án:
\(\begin{array}{l}
a)1\\
b)0\\
c)2020
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){\left[ {\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)} \right]^2}\\
= {\left( {3 - 4} \right)^2} = 1\\
b)2 - 2\sqrt 2 + 1 - \dfrac{3}{2}.2 + \dfrac{{4\sqrt 2 }}{5} + \sqrt {\dfrac{{36}}{{25}}} .\sqrt 2 \\
= 3 - 2\sqrt 2 - 3 + \dfrac{{4\sqrt 2 }}{5} + \dfrac{{6\sqrt 2 }}{5}\\
= - 2\sqrt 2 + 2\sqrt 2 = 0\\
c)\left( {1 + \sqrt {2021} } \right).\sqrt {2021 - 2.\sqrt {2021} .1 + 1} \\
= \left( {1 + \sqrt {2021} } \right).\sqrt {{{\left( {\sqrt {2021} - 1} \right)}^2}} \\
= \left( {1 + \sqrt {2021} } \right).\left( {\sqrt {2021} - 1} \right)\\
= 2021 - 1 = 2020
\end{array}\)
