`a, 5= sqrt25`
Mà `sqrt25 < sqrt27 => 5 < sqrt27`
`b, sqrt{2+1}=sqrt{3}`
Vì `3>sqrt{3}=> 3>sqrt{2+1}`
`c, sqrt{15-1}=sqrt{14}`
`3=sqrt{9}`
Vì `sqrt{9} < sqrt{14} => 3<sqrt{15-1}`
`d, 2+sqrt{6} < 2 +sqrt{9} = 2+3=5`
\(5=\sqrt{25}< 27\)
\(3>\sqrt{2+1}=\sqrt{3}\)
\(3=\sqrt{9}< \sqrt{14}\)
\(2+\sqrt{6}< 5=2+\sqrt{9}\)