So sánh A=10^2012 + 1 / 10^2011 + 1 và B =10^2011 + 1 / 20^2010 + 1
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NA
1
NB
9 tháng 3 2015
Cho C=\(10^{2010}+\frac{1}{10^{2010}}\)
Xét \(A_1=10^{2010}+\frac{1}{10^{2011}}\)và \(B^{ }_1=10^{2011}+\frac{1}{10^{2012}}\)
Ta có \(A_1-C=10^{2010}+\frac{1}{10^{2010}}-10^{2010}-\frac{1}{10^{2010}}\)
\(A_1-C=10.\left(\frac{1}{10^{2011}}-\frac{1}{10^{2010}}\right)\)
Giair tượng tự ta được \(B_1-C=10^{2010}.\left(9+\frac{1}{10^{2012}}-\frac{1}{10^{2010}}\right)\)
Ta thấy \(\frac{1}{10^{2012}}-\frac{1}{10^{2010}}
PT
2
S
2 tháng 3 2017
Vì \(\frac{10^{2011}+1}{10^{2012}+1}< 1\)
=> \(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}=\frac{10^{2011}+10}{10^{2012}+10}=\frac{10\left(10^{2010}+1\right)}{10\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy A > B
10 tháng 3 2016
Dễ thấy B < 1 vì 102011 + 1 < 102012 + 1. Áp dụng tính chất nếu \(\frac{a}{b}<1\) thì \(\frac{a}{b}<\frac{a+m}{b+m}\) ta có :
\(B=\frac{10^{2011}+1}{10^{2012}+1}<\frac{\left(10^{2011}+1\right)+9}{\left(10^{2012}+1\right)+9}=\frac{10^{2011}+10}{10^{2012}+10}=\frac{10.\left(10^{2010}+1\right)}{10.\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy A > B
30 tháng 6 2018
a) Ta có :
\(A=\frac{10^{2010}+1}{10^{2011}+1}\)
\(\Rightarrow10A=\frac{10^{2011}+10}{10^{2011}+1}=\frac{\left(10^{2011}+1\right)+9}{10^{2011}+1}=1+\frac{9}{10^{2011}+1}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}\)
\(\Rightarrow10B=\frac{10^{2012}+10}{10^{2012}+1}=\frac{\left(10^{2012}+1\right)+9}{10^{2012}+1}=1+\frac{9}{10^{2012}+1}\)
Vì \(\frac{9}{10^{2011}+1}>\frac{9}{10^{2012}+1}\)nên \(10A>10B\)
\(\Rightarrow A>B\)
Vậy : \(A>B\)
b) Ta có :
\(\left(\frac{-1}{2}\right)^{11}=\frac{-1^{11}}{2^{11}}=\frac{-1}{2^{11}}\)
\(\left(\frac{-1}{2}\right)^{13}=\frac{-1^{13}}{2^{13}}=\frac{-1}{2^{13}}\)
Vì \(\frac{-1}{2^{11}}>\frac{-1}{2^{13}}\)nên \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
Vậy : \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
N
30 tháng 6 2018
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+10}{10^{2012}+10}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10\cdot\left(10^{2010}+1\right)}{10\cdot\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy : B < A
TP
1
9 tháng 2 2018
a/ Áp dụng bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(\dfrac{10^{2011}+1}{10^{2012}+1}< 1\)
\(\Leftrightarrow\dfrac{10^{2011}+1}{10^{2012}+1}< \dfrac{10^{2011}+1+9}{10^{2012}+1+9}=\dfrac{10^{2011}+10}{10^{2012}+10}=\dfrac{10\left(10^{2010}+1\right)}{10\left(10^{2011}+1\right)}=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{10^{2011}+1}{10^{2012}+1}< \dfrac{10^{2010}+1}{10^{2011}+1}\)
S
7 tháng 5 2017
a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)
\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)
Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)
b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)
Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)
g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)
\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)
h, Vì E < 1 nên:
\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)
Vậy E = F
\(A=\dfrac{10^{2012}+1}{10^{2011}+1}\)
Mà ta có: \(10^{2012}+1>10^{2011}+1\)
\(\Rightarrow A=\dfrac{10^{2022}+1}{10^{2011}+1}>1\) (1)
\(B=\dfrac{10^{2011}+1}{20^{2010}+1}\)
Mà ta có: \(20^{2010}+1>10^{2011}+1\)
\(\Rightarrow B=\dfrac{10^{2011}+1}{20^{2010}+1}< 1\) (2)
Từ (1) và (2) \(\Rightarrow A>B\)