Không dùng máy tính, hãy so sánh A và B
A=\(\frac{352}{353}_{ }\)\(+\frac{353}{354}+\frac{354}{255}\); B=\(\frac{352+353+354}{353+354+355}\)
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\(A=355+\frac{354}{2}+\frac{353}{3}+...+\frac{2}{354}+\frac{1}{355}\)
\(A=1+\left(\frac{354}{2}+1\right)+...+\left(\frac{2}{354}+1\right)+\left(\frac{1}{355}+1\right)\)
\(A=1+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)
\(A=\frac{356}{356}+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)
\(A=356.\left(\frac{1}{2}+...+\frac{1}{354}+\frac{1}{355}+\frac{1}{356}\right)\)
Sorry , mk biết làm đến bước đấy thôi
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 }
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 ...
\(\frac{31}{61}>\frac{311}{611}\)
VÌ 61>31 LÀ 30 ĐƠN VỊ.
MÀ 611>311 LÀ 300 ĐƠN VỊ
Ta có:\(\frac{31}{61}=\frac{310}{610}\)
\(\forall a,b\in Z;b\ne0;a< b\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)
\(\Rightarrow\frac{310}{610}< \frac{310+1}{610+1}=\frac{311}{611}\)
\(\Rightarrow\frac{31}{61}< \frac{311}{611}\)
\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)
\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)
=> 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 \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)
\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)
Bài này được cái dễ lộn số =.=
Ta có :
\(B=\frac{352+353+354}{353+354+355}=\frac{352}{343+354+355}+\frac{353}{353+354+355}+\frac{354}{353+354+355}\)
Vì :
\(\frac{352}{343}>\frac{352}{353+354+355}\)
\(\frac{353}{354}>\frac{353}{353+354+355}\)
\(\frac{354}{355}>\frac{354}{353+354+355}\)
Nên \(\frac{352}{353}+\frac{353}{354}+\frac{354}{355}>\frac{352+353+354}{353+354+355}\)
Hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có : \(B=\frac{352+353+354}{353+354+355}\)
\(\Rightarrow B=\frac{352}{353+354+355}+\frac{353}{353+354+355}+\frac{354}{353+354+355}\)
Ta có : \(\frac{352}{353}>\frac{352}{353+354+355}\)
\(\frac{353}{354}>\frac{353}{353+354+355}\)
\(\frac{354}{355}>\frac{354}{353+354+355}\)
Cộng vế theo vế, ta có : \(\frac{352}{353}+\frac{353}{354}+\frac{354}{355}>\frac{352+353+354}{353+354+355}\)
\(\Rightarrow A>B\)