\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

a)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11

`@` `\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

a)\(1+\sqrt{3}>1+\sqrt{1}=1+1=2\)

Vậy \(1+\sqrt{3}>2\)

c) \(\sqrt{3}-1< \sqrt{4}-1=2-1=1\)

Vậy \(\sqrt{3}-1< 1\)

e) \(\sqrt{2}+\sqrt{5}< \sqrt{16}+\sqrt{16}=4+4=8\)

Vậy \(\sqrt{2}+\sqrt{5}< 8\)

Mình chọn nhầm lớp 8 chứ thật ra câu hỏi ở bên lớp 9 

a) Ta có \(5=\sqrt{25}\)

Vì \(\sqrt{25}>\sqrt{11}\) nên \(5>\sqrt{11}\)

b) Ta có \(4=\sqrt{16}\)

Vì \(\sqrt{13}< \sqrt{16}\) nên \(\sqrt{13}< 4\)

c) Ta có \(-7=-\sqrt{49}\)

Vì \(-\sqrt{49}< -\sqrt{43}\) nên \(-7< -\sqrt{43}\)

d) Ta có \(-5=-\sqrt{25}\)

Vì \(-\sqrt{21}>-\sqrt{25}\) nên \(-\sqrt{21}>-5\)

\(\left(\sqrt{8}+\sqrt{11}\right)^2=8+2\sqrt{8}.\sqrt{11}+11=19+\sqrt{352}\)\(< 19+\sqrt{361}=19+19=38=\left(\sqrt{38}\right)^2\)

\(\Leftrightarrow\left(\sqrt{8}+\sqrt{11}\right)^2< \left(\sqrt{38}\right)^2\Rightarrow\sqrt{8}+\sqrt{11}< \sqrt{38}\)

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a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)