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311<411; 411=(22)11=222
1714>1614; 1614=(24)14=256
Vì 222<256=> 411<1614
=>311<1714
![](https://rs.olm.vn/images/avt/0.png?1311)
Giả sử \(5< 2+\sqrt[]{11}\)
\(\Leftrightarrow3< \sqrt[]{11}\)
\(\Leftrightarrow9< 11\left(luôn.đúng\right)\)
Vậy \(5< 2+\sqrt[]{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\sqrt{3}+5> \sqrt{1}+5=6$
$\sqrt{2}+\sqrt{11}< \sqrt{4}+\sqrt{16}=6$
$\Rightarrow \sqrt{3}+5> \sqrt{2}+\sqrt{11}$
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`\sqrt {2} + \sqrt {11}` và `\sqrt {3} + 5`
Ta có: `5^2 = 25`
`=> \sqrt {25} = 5`
`=> \sqrt {3} + 5 = \sqrt {3} + \sqrt {25}`
Vì: \(\left\{{}\begin{matrix}\sqrt{3}>\sqrt{2}\\\sqrt{25}>\sqrt{11}\end{matrix}\right.\)
`=>`\(\sqrt{3}+\sqrt{25}>\sqrt{2}+\sqrt{11}\)
`=> \sqrt {3} + 5 > \sqrt {2} + \sqrt {11}.`
`# \text {NgMH}`
(căn 2+căn 11)^2=13+2*căn 22
(căn 3+5)^2=28+2*căn 45
mà 13<28; căn 22<căn 45
nên căn 2+căn 11<căn 3+5
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: \(\sqrt{3}-\sqrt{5}< 0\)
nên \(\sqrt{3}-\sqrt{5}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)
mà 80>75
nên \(4\sqrt{5}>5\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: căn 14<4
=>7+căn 14<4+7=11
b: -căn 5<-2
=>-căn 5+9<-2+9=7
d: \(\sqrt{145}< 13\)
=>-11+căn 145<-11+13=2
e: \(7-4\sqrt{5}+2=9-4\sqrt{5}>0\)
=>7-4căn 5>-2
f: -4căn 5>-9
=>-9-4căn 5>-9-9=-18
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8^2=64=32+32\\ \left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{255}\)
\(32^2=1024>1020=\left(2\sqrt{255}\right)^2\\ \Rightarrow64>32+2\sqrt{255}\\ \Rightarrow8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\\ \Leftrightarrow8>\sqrt{15}+\sqrt{17}\\ \Leftrightarrow-8< -\left(\sqrt{15}+\sqrt{17}\right)\)
Lời giải:
\((\sqrt{15}+\sqrt{17})^2=32+2\sqrt{15.17}=32+2\sqrt{(16-1)(16+1)}\)
\(=32+2\sqrt{16^2-1}< 32+2\sqrt{16^2}=64\)
\(\Rightarrow \sqrt{15}+\sqrt{17}< 8\)
\(\Rightarrow -(\sqrt{15}+\sqrt{17})> -8\)
Giả sử \(-5\sqrt{11}< -17\)
\(\Leftrightarrow 17< 5\sqrt{11} \Leftrightarrow \left(17\right)^2< \left(5\sqrt{11}\right)^2\)
\(\Leftrightarrow289< 25.11\Leftrightarrow289< 275\left(sai\right)\)
vậy \(-5\sqrt{11}>-17\)