a) Ta có: \(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^2-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^4-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^8-1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{16}-1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)
\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{32}-1\right]\)
\(=65535\sqrt{2}+65535\)
b) Ta có: \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)
\(=2\sqrt{505}-1\)
c) Ta có: \(C^3=26+15\sqrt{3}+26-15\sqrt{3}+3\cdot\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\cdot\left(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\right)\)
\(\Leftrightarrow C^3=52+3\cdot C\)
\(\Leftrightarrow C^3-3\cdot C-52=0\)
\(\Leftrightarrow C^3-4C^2+4C^2-16C+13C-52=0\)
\(\Leftrightarrow C^2\left(C-4\right)+4C\left(C-4\right)+13\left(C-4\right)=0\)
\(\Leftrightarrow\left(C-4\right)\left(C^2+4C+13\right)=0\)
mà \(C^2+4C+13>0\)
nên C-4=0
hay C=4