ta có:

\(A^2=\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}+2\sqrt{\dfrac{\left(3\sqrt{3}-4\right)\left(\sqrt{3}+4\right)}{\left(2\sqrt{3}+1\right)\left(5-2\sqrt{3}\right)}}+\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}\)

\(A^2=\dfrac{\left(3\sqrt{3}-4\right)\left(5-2\sqrt{3}\right)+\left(2\sqrt{3}+1\right)\left(\sqrt{3}+4\right)}{\left(2\sqrt{3}+1\right)\left(5-2\sqrt{3}\right)}+2\sqrt{\dfrac{9+12\sqrt{3}-4\sqrt{3}-16}{10\sqrt{3}-12+5-2\sqrt{3}}}\)

\(A^2=\dfrac{15\sqrt{3}-18-20+8\sqrt{3}+6+8\sqrt{3}+\sqrt{3}+4}{10\sqrt{3}-12+5-2\sqrt{3}}+2\sqrt{\dfrac{8\sqrt{3}-7}{8\sqrt{3}-7}}\)

\(A^2=\dfrac{32\sqrt{3}-28}{8\sqrt{3}-7}+2\)

\(A^2=6\Rightarrow a=\sqrt{6}\)

ta có \(2A=\left(x^2+2xy+y^2\right)+\left(x^2-6x+9\right)+\left(y^2-6x+9\right)+4032\)

\(2A=\left(x+y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2+4032\)

\(\Rightarrow2A\ge4032\Leftrightarrow A\ge2016\)

A = 1 + 3² + 3⁴ + ...+ 3²⁰⁰²

A = ( 1 + 3² + 3⁴ ) + ( 3⁶ + 3⁸ + 3¹⁰ ) + ... + ( 3¹⁹⁹⁸ + 3²⁰⁰⁰ + 3²⁰⁰² )

A = 91 + 3⁶(1+3²+3⁴) + ... + 3¹⁹⁹⁸(1+3²+3⁴)

A = 91.(3⁶ + 3⁸ + ... + 3¹⁹⁹⁸ ) chia hết cho 7 

=> đpcm