`a)` Ta có:
`\qquad 2\sqrt{27}=\sqrt{2^2 .27}=\sqrt{108}`
Vì `108<147=>\sqrt{108}<\sqrt{147}`
`=>2\sqrt{27}<\sqrt{147}`
$\\$
`b)` Ta có:
`\qquad -3\sqrt{5}=-\sqrt{3^2 . 5}=-\sqrt{45}`
`\qquad -5\sqrt{3}=-\sqrt{5^2 . 3}=-\sqrt{75}`
Vì `45<75=>\sqrt{45}<\sqrt{75}`
`=>-\sqrt{45}> -\sqrt{75}`
`=>-3\sqrt{5}> -5\sqrt{3}`
$\\$
`c)` Ta có:
`-\sqrt{123}<0` và `21>0;2\sqrt{7}>0;15\sqrt{3}>0`
`=>-\ sqrt{123}` bé nhất
`\qquad 21=\sqrt{21^2}=\sqrt{441}`
`\qquad 2\sqrt{7}=\sqrt{2^2 . 7}=\sqrt{28}`
`\qquad 15\sqrt{3}=\sqrt{15^2 . 3}=\sqrt{675}`
Vì `0<28<441<675`
`=>\sqrt{28}<\sqrt{441}<\sqrt{675}`
`=>2\sqrt{7}<21<15\sqrt{3}`
Vậy thứ tự tăng dần của các số đã cho là:
`-\sqrt{123}; 2\sqrt{7};21;15\sqrt{3}`
$\\$
`d)` Ta có:
`\qquad 2\sqrt{15}=\sqrt{2^2 .15}=\sqrt{60}`
Vì `60>59>0=>\sqrt{60}>\sqrt{59}`
`=>2\sqrt{15}>\sqrt{59}`
$\\$
`e)` Ta có:
`\qquad 2\sqrt{2}=\sqrt{2^2 .2}=\sqrt{8}`
Vì `0<8<9=>\sqrt{8}<\sqrt{9}`
`=>2\sqrt{2}<\sqrt{3^2}=3`
`=>2\sqrt{2}-1<3-1`
`=>2\sqrt{2}-1<2`
$\\$
`f)` Ta có: `0<36<41`
`=>\sqrt{36}<\sqrt{41}`
`=>\sqrt{6^2}<\sqrt{41}`
`=>6<\sqrt{41}`
Vậy: `6<\sqrt{41}`
$\\$
`g)` Ta có:
`\qquad 0<3<4=>\sqrt{3}<\sqrt{4}`
`=>\sqrt{3}<2`
`=>\sqrt{3}/2<2/2`
`=>\sqrt{3}/2<1`
$\\$
`h)` Ta có:
`\qquad 0<10<20=>\sqrt{10}<\sqrt{20}`
`=>\sqrt{10}<\sqrt{2^2 .5}`
`=>\sqrt{10}<2\sqrt{5}`
`=>\sqrt{10}/2< \sqrt{5}<2\sqrt{5}`
`=>-\sqrt{10}/2> -2\sqrt{5}`
$\\$
`i)` Ta có: `0<6<16=>\sqrt{6}<\sqrt{16}`
`=>\sqrt{6}<\sqrt{4^2}`
`=>\sqrt{6}<4`
`=>\sqrt{6}-1<4-1`
`=>\sqrt{6}-1<3`
$\\$
`j)` Ta có:
`2\sqrt{5}=\sqrt{2^2 .5}=\sqrt{20}`
`5\sqrt{2}=\sqrt{5^2 . 2}=\sqrt{50}`
Vì `0<20<50=>\sqrt{20}<\sqrt{50}`
`=>2\sqrt{5}<5\sqrt{2}`
`=>2\sqrt{5}-5\sqrt{2}<0<1`
Vậy: `2\sqrt{5}-5\sqrt{2}<1`
$\\$
`k)` Ta có:
`\qquad 4\sqrt{8}=\sqrt{4^2 .8}=\sqrt{128}`
`\qquad 9=\sqrt{9^2}=\sqrt{81}`
Vì `128>81>0`
`=>\sqrt{128}>\sqrt{81}`
`=>4\sqrt{8}>9`
`=>{4\sqrt{8}}/{3.4}> 9/{3.4}`
`=>\sqrt{8}/3>3/4`
