a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

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) Ta có: 1,(32) = 1,323232….

Quan sát chữ số ở hàng thập phân thứ 2, ta thấy 1 < 2 nên 1,313233… < 1,(32)

b) Ta có: \(\sqrt 5  = 2,236 \ldots .\)

Quan sát chữ số ở hàng thập phân thứ nhất, ta thấy 2 < 3 nên 2,236 < 2,36

Vậy \(\sqrt 5 \) < 2,36

\(2\sqrt{3+\sqrt{5}}=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{2}\cdot\left(\sqrt{5}+1\right)\)

\(=\sqrt{10}+\sqrt{2}>\sqrt{10}+1\)

Vậy ....

a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)

\(\sqrt{6}< \sqrt{6,25}=2,5\); 

\(\sqrt{12}< \sqrt{12,25}=3,5\); 

\(\sqrt{20}< \sqrt{20,25}=4,5\)

=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)

Vậy P < 12

Answer:

ý a, tham khảo bài làm của @xyzquynhdi

\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)

\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

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

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

\(P=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

\(Q=\dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)

Do \(2< \sqrt{2}+1\)

=> P < Q

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 }