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)2=1+1

Có:1²<\(\sqrt{2}^{^{ }2}\)

=> 1<\(\sqrt{2}\)

=>1+1<\(\sqrt{2}+1\)

=>2<\(\sqrt{2}+1\)

c) 10=2.5

Có;\(5=\)\(\sqrt{25}< \sqrt{31}\)

=>\(\sqrt{31}>\sqrt{25}\)

=>\(2.\sqrt{31}>2.\sqrt{25}\)

=>\(2.\sqrt{31}>10\)

b) 1=2-1

Có: \(2=\sqrt{4}>\sqrt{3}\)

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

=>\(1>\sqrt{3}-1\)

d) -12=-3.4

Có:\(4=\sqrt{16}>\sqrt{11}\)

=>\(\sqrt{11}< \sqrt{16}\)

=>\(-3.\sqrt{11}>-3.\sqrt{16}\)

=>\(-3.\sqrt{11}>-12\)

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

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

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

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) \(1=\sqrt{1}< \sqrt{2}\)

b) \(2=\sqrt{4}>\sqrt{3}\)

c) \(6=\sqrt{36}< \sqrt{41}\)

d) \(7=\sqrt{49}>\sqrt{47}\)

e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)

f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)

g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)

h) \(\sqrt{3}>0>-\sqrt{12}\)

i) \(5=\sqrt{25}< \sqrt{29}\)

\(\Rightarrow-5>-\sqrt{29}\)