EC
2
K
Khách
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24 tháng 9 2016
a) Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2009^{2009}+1+2007}\)
\(A< \frac{2008^{2008}+2008}{2008^{2009}+2008}\)
\(A< \frac{2008.\left(2008^{2007}+1\right)}{2008.\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)
=> A < B
b) Áp dụng \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)
\(N>\frac{100^{101}+100}{100^{100}+100}\)
\(N>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)
=> M > N
10 tháng 7 2019
1
\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)
\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)
10 tháng 7 2019
2
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
NM
1
H9
HT.Phong (9A5)
CTVHS
21 tháng 9 2023
Ta có:
\(M=\dfrac{100^{100}+1}{100^{99}+1}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)
\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\)
\(N=\dfrac{100^{101}+1}{100^{100}+1}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)
\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)
Mà: \(100^{101}>100^{100}\)
\(\Rightarrow100^{101}+100>100^{100}+100\)
\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)
\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)
\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)
\(\Rightarrow N< M\)
EC
1
21 tháng 10 2015
Có :
A=100100+1/10099+1
1/100.A=100100+1/100.(10099+1)
A/100=100100+1/100100+100
A/100=1-99/100100
B bạn cũng làm tương tự và sau đó bạn so sánh 99/100^100 Và 99/100^69 là Ok.
NL
1
9 tháng 7 2016
Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)
Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)
\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)
9 tháng 7 2017
a, A = \(\frac{1}{2}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)
\(A=\frac{1}{2}.\left(\frac{3.4....99}{4.5...100}\right)\)
\(A=\frac{1}{2}.\left(\frac{3}{100}\right)\)\(\)\(A=\frac{3}{200}\)
\(B=\frac{2}{3}.\frac{4}{5}.\frac{5}{6}...\frac{100}{101}\)
\(B=\frac{2}{3}.\left(\frac{4.5...100}{5.6...101}\right)\)
\(B=\frac{2}{3}.\left(\frac{4}{101}\right)\)
\(B=\frac{8}{303}\)
\(A.B=\frac{8}{303}.\frac{3}{200}\)
\(A.B=\frac{1}{2525}\)
b, A = 1/2 x 3/100
B = 2/3 x 4/101
Ta có : 1 - 2/3 = 1/3; 1 - 1/2 = 1/2
MÀ 1/3 < 1/2 => 2/3 > 1/2 (1)
Ta có : 1 - 3/100 = 97/100
1 - 4/101 = 97/101
Mà 97/101 < 97/100 => 4/101 > 3/100 (2)
Từ (1) và (2) => B > A
9 tháng 7 2017
a,
\(AB=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(AB=\frac{\left[1\cdot3\cdot5\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
b,
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\Leftrightarrow A< B\)
M= \(\frac{100^{100}+1}{100^{99}+1}=\frac{100^{100}+100-99}{100^{99}+1}=\frac{100^{100}+100}{100^{99}+1}-\frac{99}{100^{99}+1}=\frac{100.\left(100^{99}+1\right)}{100^{99}+1}-\frac{99}{100^{99}+1}\)
\(=100-\frac{99}{100^{99}+1}\)
N= \(\frac{100^{101}+1}{100^{100}+1}=\frac{100^{101}+100-99}{100^{100}+1}=\frac{100^{101}+100}{100^{100}+1}-\frac{99}{100^{100}+1}\)
\(=\frac{100.\left(100^{100}+1\right)}{100^{100}+1}-\frac{99}{100^{100}+1}=100-\frac{99}{100^{100}+1}\)
Vi 100100+1>10099+1
=> \(\frac{99}{100^{99}+1}>\frac{99}{100^{100}+1}\)
=> \(100-\frac{99}{100^{99}+1}
uk ai cũng có lúc nhầm mà chẳng sao đâu bạn ak