so sánh \(\frac{-1}{\sqrt{33}-\sqrt{31}}\) và \(\frac{-1}{\sqrt{34}-\sqrt{32}}\)
K
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NP
0
LC
20 tháng 8 2015
Đặt \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}\)
Ta thấy: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2015}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{2015}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{2015}}\)
.........................
\(\frac{1}{\sqrt{2014}}>\frac{1}{\sqrt{2015}}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2014}}>\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+...+\frac{1}{\sqrt{2015}}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2014}}+\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+...+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}\)
=>\(A>2015.\frac{1}{\sqrt{2015}}=\frac{2015}{\sqrt{2015}}=\sqrt{2015}\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}>\sqrt{2015}\)
QL
1
KN
8 tháng 7 2019
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
..........
..........
..........
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100=\frac{100}{10}=10\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
22 tháng 7 2018
Ta có
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
........................................
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)(100 số\(\frac{1}{10}\)) >10
TN
2 tháng 7 2017
B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)
\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)
\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)
\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))
FF
0
MH
9 tháng 6 2015
Hình như bạn hơi nhầm đề bài . Nếu B là 10 thì mình biết .
Nhận thấy : \(\frac{1}{\sqrt{1}}\)>\(\frac{1}{\sqrt{100}}\); \(\frac{1}{\sqrt{2}}\)>\(\frac{1}{\sqrt{100}}\);...: \(\frac{1}{\sqrt{100}}\)=\(\frac{1}{\sqrt{100}}\)
<=> A= \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+\(\frac{1}{\sqrt{3}}\)+...+\(\frac{1}{\sqrt{100}}\)>\(\frac{1}{\sqrt{100}}\)+\(\frac{1}{\sqrt{100}}\)+...+\(\frac{1}{\sqrt{100}}\)( 100 số \(\frac{1}{\sqrt{100}}\))
Hay : A > \(\frac{1}{\sqrt{100}}\).100
<=> A > 10
<=> A>B
Nếu không đúng mong bạn thông cảm nhé !!
LN
8 tháng 12 2017
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 }
+ \(-\frac{1}{\sqrt{33}-\sqrt{31}}=-\frac{\sqrt{33}+\sqrt{31}}{\left(\sqrt{33}-\sqrt{31}\right)\left(\sqrt{33}+\sqrt{31}\right)}\)
\(=-\frac{\sqrt{33}+\sqrt{31}}{2}\)
+ \(-\frac{1}{\sqrt{34}-\sqrt{32}}=-\frac{\sqrt{34}+\sqrt{32}}{2}\)
+ \(\sqrt{34}+\sqrt{32}>\sqrt{33}+\sqrt{31}\)
\(\Rightarrow-\left(\sqrt{34}+\sqrt{32}\right)< -\left(\sqrt{33}+\sqrt{31}\right)\)
\(\Rightarrow-\frac{\sqrt{33}+\sqrt{31}}{2}>-\frac{\sqrt{34}+\sqrt{32}}{2}\)
\(\Rightarrow-\frac{1}{\sqrt{33}-\sqrt{31}}>-\frac{1}{\sqrt{34}-\sqrt{32}}\)