a) Ta có :\((\sqrt{2}+\sqrt{3})^2=2+3+\sqrt{24}<2+3+5=10\)
\(\sqrt{10}^2=10\)
\(=>(\sqrt{2}+\sqrt{3})^2<\sqrt{10}^2\)
\(=> \sqrt{2}+\sqrt{3}<\sqrt{10}\)
b) Ta có : \((\sqrt{3}+2)^2=3+4+\sqrt{48}=7+\sqrt{48}\)
\((\sqrt{2}+\sqrt{6})^2=2+6+\sqrt{48}=8+\sqrt{48}\)
\(=>(\sqrt{3}+2)^2<(\sqrt{2}+\sqrt{6})^2\)
\(=>\sqrt{3}+2<\sqrt{2}+\sqrt{6}\)
Ta có
\(\left(\sqrt{2}+\sqrt{3}\right)^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt[]{6}\)
Ta cũng có \(\sqrt{10}^2=10\)
Nên \(\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{10}^2\\
=5+2\sqrt{6}-10\\ =2\sqrt{6}-5\\ =\sqrt{4.6}-\sqrt{25}\\ =\sqrt{24}-\sqrt{25}< 0\)
Do đó \(\sqrt{2}+\sqrt{3}< \sqrt{10}\)
b) Ta có \(\left(\sqrt{3}+2\right)^2=3+4\sqrt{3}+4=7+4\sqrt{3}\)
\(\left(\sqrt{2}+\sqrt{6}\right)^2=2+2\sqrt{12}+6=8+2\sqrt{12}\)
Do đó \(\left(\sqrt{3}+2\right)^2-\left(\sqrt{6}+\sqrt{2}\right)^2\\ =7+4\sqrt{3}-8-2\sqrt{12}\\ =-1+\sqrt{16.3}-\sqrt{4.12}\\ =-1+\sqrt{48}-\sqrt{48}=-1< 0\)
Nên \(\sqrt{3}+2< \sqrt{6}+\sqrt{2}\)