a) đặt A = \(\sqrt{14+8\sqrt{3}}.\left(2\sqrt{2}+\sqrt{3}\right)\)
=> \(A^2=\left(14+8\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(=\left(14+8\sqrt{3}\right)\left(14+8\sqrt{3}\right)\)
\(=\left(14+8\sqrt{3}\right)^2\)
=> A = \(14+8\sqrt{3}\)
b) đặt B = \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
=> \(B^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
= \(4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
= \(8-2\sqrt{9}\)
\(=8-6=2\)
=> C = \(\sqrt{2}\)
c) đặt C = \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
=> \(C^2=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+3+\sqrt{5}\)
\(=6+2\sqrt{1}\) \(=8\)
=> C = \(\sqrt{8}\)
mong bài mk đúng :)~~
