a) Ta có: \(\left(\sqrt{3}+2\right)^2=3+2\cdot\sqrt{3}\cdot2+2^2\)
\(=7+4\sqrt{3}\)
Ta có: \(\left(\sqrt{2}+\sqrt{6}\right)^2=2+2\cdot\sqrt{2}\cdot\sqrt{6}+6\)
\(=8+4\sqrt{3}\)
mà \(7+4\sqrt{3}< 8+4\sqrt{3}\)
nên \(\left(\sqrt{3}+2\right)^2< \left(\sqrt{2}+\sqrt{6}\right)^2\)
hay \(\sqrt{3}+2< \sqrt{2}+\sqrt{6}\)
b) Ta có: \(16=\sqrt{256}\)
\(\sqrt{15}\cdot\sqrt{17}=\sqrt{15\cdot17}=\sqrt{255}\)
mà \(\sqrt{256}>\sqrt{255}\)
nên \(16>\sqrt{15}\cdot\sqrt{17}\)
c) Ta có: \(\left(6+2\sqrt{2}\right)^2=36+2\cdot6\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2\)
\(=36+24\sqrt{2}+8\)
\(=44+24\sqrt{2}=44+\sqrt{1152}\)
\(9^2=81=44+37=44+\sqrt{1369}\)
mà \(44+\sqrt{1152}< 44+\sqrt{1369}\)
nên \(\left(6+2\sqrt{2}\right)^2< 9^2\)
hay \(6+2\sqrt{2}< 9\)
d) Ta có: \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)
\(3^2=9=5+4=5+\sqrt{16}\)
mà \(5+\sqrt{24}>5+\sqrt{16}\)
nên \(\left(\sqrt{2}+\sqrt{3}\right)^2>3^2\)
hay \(\sqrt{2}+\sqrt{3}>3\)
