so sánh m và n biết M=100^100+1/100^99+1 N=100^101+1/100^100+1
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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: \(100^{101}>100^{100}\)
\(\Rightarrow100^{101}+100>100^{100}+100\)
\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)
\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)
\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)
\(\Rightarrow N< M\)
20 tháng 10 2015
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
=> \(\frac{99}{100^{99}+1}>\frac{99}{100^{100}+1}\)
=> \(100-\frac{99}{100^{99}+1}<100-\frac{99}{100^{100}+1}\)
=> M<N
8 tháng 7 2016
Ta có:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
\(...\)
\(\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow M< N\)
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
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
12 tháng 7 2016
Áp dụng a/b > 1 => a/b > a+m/b+m (a;b;m thuộc N*)
=> \(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)
\(>\frac{100^{101}+100}{100^{100}+100}\)
\(>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)
=> N > M
Ủng hộ mk nha ^_-
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
22 tháng 2 2018
Ta có : N = \(\frac{100^{101}+1}{100^{100}+1}\)< \(\frac{100^{101}+1+99}{100^{100}+1+99}\)= \(\frac{100^{101}+100}{100^{100}+100}\)= \(\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)= \(\frac{100^{100}+1}{100^{99}+1}\)= M
Vậy M > N.
NHỚ K VỚI NHÉ!!!!!!
VT
4
1 tháng 8 2016
(10099+9999)100=10099x100+9999x100
(100100+99100)99=100100x99+99100x99
Vì100100x99+99100x99=10099x100+9999x100
=>M=N
Các bạn nhớ nha !!!
PD
0
\(M=\frac{100^{100}+1}{100^{99}+1}=\frac{100^{100}+100-99}{100^{99}+1}=100-\frac{99}{100^{99}+1}\)
\(N=\frac{100^{101}+1}{100^{100}+1}=\frac{100^{101}+100-99}{100^{100}+1}=100-\frac{99}{100^{100}+1}\)
\(\frac{99}{100^{99}+1}>\frac{99}{100^{100}+1}\Rightarrow A< B\)