\(A=355+\frac{354}{2}+\frac{353}{3}+...+\frac{2}{354}+\frac{1}{355}\)

\(A=1+\left(\frac{354}{2}+1\right)+...+\left(\frac{2}{354}+1\right)+\left(\frac{1}{355}+1\right)\)

\(A=1+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)

\(A=\frac{356}{356}+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)

\(A=356.\left(\frac{1}{2}+...+\frac{1}{354}+\frac{1}{355}+\frac{1}{356}\right)\)

Sorry , mk biết làm đến bước đấy thôi 

cảm ơn bạn

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 }

tôi thích hoa hồng sai kìa

Vì 2006/2007 ; 2007/2008 ; 2008/2009 ; 2009/2010 đều bé hơn 1 nên:

2006/2007 + 2007/2008 + 2008/2009 + 2009/2010 < 1 + 1 + 1 + 1 = 4.

Vậy ...

A=\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)=3-(\(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\))+1+\(\frac{3}{2006}\)=4+(\(\frac{1}{2006}-\frac{1}{2007}\))+(\(\frac{1}{2006}-\frac{1}{2008}+\frac{1}{2009}\))

=> A>4 (\(\frac{1}{2006}>\frac{1}{2007}>\frac{1}{2008}>\frac{1}{2009}\))

\(\frac{31}{61}>\frac{311}{611}\)

VÌ 61>31  LÀ 30 ĐƠN VỊ.

MÀ 611>311 LÀ 300 ĐƠN VỊ

Ta có:\(\frac{31}{61}=\frac{310}{610}\)

\(\forall a,b\in Z;b\ne0;a< b\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

\(\Rightarrow\frac{310}{610}< \frac{310+1}{610+1}=\frac{311}{611}\)

\(\Rightarrow\frac{31}{61}< \frac{311}{611}\)

gjugujk

Có Ta có\(VT=\frac{2014}{\sqrt{2015}}+\frac{2015}{\sqrt{2014}}=\frac{2015-1}{\sqrt{2015}}+\frac{2014+1}{\sqrt{2014}}=\sqrt{2015}-\frac{1}{\sqrt{2015}}+\sqrt{2014}+\frac{1}{\sqrt{2014}}.\)​​​\(20140\Leftrightarrow VT>VP\)

Cau thu lay 405:8 va 354:7 xem cai nao lon hon h cho to nhe

\(\frac{354}{7}\)lớn hơn \(\frac{405}{8}\)

\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)

\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)

=> Bằng nhau

\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)

\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)

vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)