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

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

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

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

a/ \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2.3}=5+2\sqrt{6}=5+\sqrt{24}\)

\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)

Vì \(\sqrt{24}< \sqrt{25}\)

=>\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)

b/\(\left(\sqrt{3}+2\right)^2=3+4+4\sqrt{3}=7+4\sqrt{3}\)

\(\left(\sqrt{2}+\sqrt{16}\right)^2=2+16+2\sqrt{2.16}=18+4\sqrt{8}\)

=> \(\sqrt{3}+2< \sqrt{2}+\sqrt{16}\)

c/ \(16=\sqrt{16^2}\)

\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)

=> \(16>\sqrt{15}.\sqrt{17}\)

d/\(8^2=64=32+32=32+2\sqrt{256}\)

\(\left(\sqrt{15}+\sqrt{17}\right)^2=15+17+2\sqrt{15.17}=32+2\sqrt{255}\)

=> \(8>\sqrt{15}+\sqrt{17}\)

a. Ta có : \(\sqrt{8}< \sqrt{9}\) ( vì 8< 9)

hay \(2\sqrt{2}< 3\)

\(\Rightarrow\) \(2\sqrt{2}+6< 3+6\)

hay \(2\sqrt{2}+6< 9\)

b. Ta có : \(\sqrt{6}>\sqrt{4}\) (vì 6 > 4 )

hay \(\sqrt{2.3}>2\)

\(\Rightarrow\) 2\(\sqrt{2.3}\) > 4

\(\Rightarrow\) 2 + \(2\sqrt{2.3}\) + 3 > 9

hay \(\left(\sqrt{2}+\sqrt{3}\right)^2\)> 9

\(\Rightarrow\) \(\sqrt{2}+\sqrt{3}>3\)

c. Ta có: \(\sqrt{80}>\sqrt{49}\) (vì 80>49)

hay \(4\sqrt{5}\) > 7

\(\Rightarrow\) 9 + \(4\sqrt{5}\) > 16

d. Ta có : \(2\sqrt{33}>2\sqrt{25}\) (vì 33> 25 ) hay \(2\sqrt{23}>2.5\)

\(\Rightarrow\) - \(2\sqrt{33}\) < - 2.5

\(\Rightarrow\) 11 - \(2\sqrt{11.3}\) +3 < 11- 2.5 +3

hay \(\left(\sqrt{11}-\sqrt{3}\right)^2\) < 4

\(\Rightarrow\) \(\sqrt{11}-\sqrt{3}< 2\)

a,  \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\Rightarrow1+1< \sqrt{2}+1\Rightarrow2< \sqrt{2}+1\)

c, \(4>3=>\sqrt{4}>\sqrt{3}=>\sqrt{4}-1>\sqrt{3}-1\Rightarrow1>\sqrt{3}-1\)

d, \(16>11=>\sqrt{16}>\sqrt{11}\Rightarrow4>\sqrt{11}=>4.\left(-3\right)< \sqrt{11}.\left(-3\right)\)

\(=>-12< -3.\sqrt{11}\) 

Lời giải:

a.

$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$

$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.

$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$

$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$

$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$

\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)

\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

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 \(4\sqrt{5}>5\sqrt{3}\)

a) \(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{2^3.3}=\sqrt[3]{24}\)

Ta có : \(24>23\), nên \(\sqrt[3]{24}>\sqrt[3]{23}\)

Vậy \(2\sqrt[3]{3}>\sqrt[3]{23}\)

b) Ta có :

\(11=\sqrt[3]{11^3}=\sqrt[3]{1331}\)

Từ đó suy ra \(33< 3\sqrt[3]{1333}\)

a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

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