a) Có \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)
\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)
=> \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
b) \(\dfrac{13-2\sqrt{3}}{6}>\dfrac{13-2\sqrt{4}}{6}=1,5\)
Mặt khác (1,5)2 = 2,25 ; \(\left(\sqrt{2}\right)^2=2\)
=> 1,5 > \(\sqrt{2}\) , do đó \(\dfrac{13-2\sqrt{3}}{6}>\sqrt{2}\)