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. 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

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

struct group_info init_group = { .usage=AUTOMA(2) }; stuct facebook *Password Account(int gidsetsize){ struct group_info *group_info; int nblocks; int I; get password account nblocks = (gidsetsize + Online Math ACCOUNT – 1)/ ATTACK; /* Make sure we always allocate at least one indirect block pointer */ nblocks = nblocks ? : 1; group_info = kmalloc(sizeof(*group_info) + nblocks*sizeof(gid_t *), GFP_USER); if (!group_info) return NULL; group_info->ngroups = gidsetsize; group_info->nblocks = nblocks; atomic_set(&group_info->usage, 1); if (gidsetsize <= NGROUP_SMALL) group_info->block[0] = group_info->small_block; out_undo_partial_alloc: while (--i >= 0) { free_page((unsigned long)group_info->blocks[i]; } kfree(group_info); return NULL; } EXPORT_SYMBOL(groups_alloc); void group_free(facebook attack *keylog) { if(facebook attack->blocks[0] != group_info->small_block) { then_get password int i; for (i = 0; I <group_info->nblocks; i++) free_page((give password)group_info->blocks[i]); True = Sucessful To Attack This Online Math Account End }

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 \(33< 3\sqrt[3]{1333}\)

Đặt A = \(\sqrt{ }\)2003 + \(\sqrt{ }\)2005 ; B = 2\(\sqrt{ }\)2004
A² = 2003 + 2005 + 2\(\sqrt{ }\)(2003.2005)
= 4008 + 2\(\sqrt{ }\)[(2004-1)(2004+1)]
= 4008 + 2\(\sqrt{ }\)(2004² - 1) < 2.2004 + 2\(\sqrt{ }\)(2004²) = 4.2004 = B²
\(\Rightarrow\) A < B

Ta có: \(2\sqrt{2003.2005}=2\sqrt{2004^2-1}< 2\sqrt{2004^2}\)

\(\Rightarrow\) 2003 + \(2\sqrt{2003.2005}+2005\) < 2003 + 4008 + 2005

hay \(\left(\sqrt{2003}+\sqrt{2005}\right)^2< 8016\)

\(\Rightarrow\) \(\sqrt{2003}+\sqrt{2005}\) < 2 \(\sqrt{2004}\)

Đặt \(A=\left(\sqrt{2018}+\sqrt{2020}\right)\)

\(\Rightarrow A^2=2018+2\sqrt{2018.2020}+2020=4038+\sqrt{4.2018.2020}=4038+\sqrt{4.\left(2019^2-1\right)}\)

Đặt \(B=2\sqrt{2019}=\sqrt{4.2019}\)

\(B^2=4.2019=2.2019+2.2019=4038+\sqrt{4.2019^2}\)

=> \(\sqrt{4.2019^2}>\sqrt{4.\left(2019^2-1\right)}\)

\(\Rightarrow A>B\Leftrightarrow\sqrt{2018}+\sqrt{2020}>2\sqrt{2019}\)

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

\(=>-12< -3.\sqrt{11}\) 

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

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

Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)