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B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)
C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
2009/2010=1-1/2010<1-1/2011=2010/2011
vậy 2009/2010<2010/2011
3^400=(3^4)^100=81^100>64^100=4^300
=>1/3^400<1/4^300
Vậy 1/3^400<1/4^300
nghịch đảo 2 phân số ta có: \(\frac{2010}{2009}v\text{à}\frac{2011}{2010}\)
phân tích ra ta có:\(\frac{2010}{2009}=1+\frac{1}{2009}\)
\(\frac{2011}{2010}=1+\frac{1}{2010}\)
Vì \(\frac{1}{2009}>\frac{1}{2010}\)
nên \(\frac{2009}{2010}
a/ Do : 2009/2010 > 2009/2011, 2009/2011 < 2010/2011 nên 2009/2010 < 2010/2011
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{20101}{2009+2010+2011}\)
Ta thấy \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\);
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Suy ra : A > B
a) Ta có:
\(1-\frac{2009}{2010}=\frac{1}{2010}\)
\(1-\frac{2010}{2011}=\frac{1}{2011}\)
Vì \(\frac{1}{2010}>\frac{1}{2011}\)=> \(\frac{2009}{2010}<\frac{2010}{2011}\)
b) Ta có:
\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)
\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)
Vì 81100 > 64100 => \(\frac{1}{81^{100}}<\frac{1}{64^{100}}\)=> \(\frac{1}{3^{400}}<\frac{1}{4^{300}}\)
c) Ta có:
\(\frac{2008}{2008\cdot2009}=\frac{1}{2009}\)
\(\frac{2009}{2009\cdot2010}=\frac{1}{2010}\)
Vì \(\frac{1}{2009}>\frac{1}{2010}\) => \(\frac{2008}{2008\cdot2009}>\frac{2009}{2009\cdot2010}\)
d) Ta có:
\(\frac{200}{201}+\frac{201}{202}=\frac{200\cdot202+201^2}{201\cdot202}>1\)
\(\frac{200+201}{201+202}=\frac{401}{403}<1\)
=> \(\frac{200\cdot202+201^2}{201\cdot202}>\frac{401}{403}\)=> \(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)
a)ta có:
\(1-\frac{2009}{2010}=\frac{1}{2010};1-\frac{2010}{2011}=\frac{1}{2011}\)
dự vào công thức so sánh phần bù
vì \(\frac{1}{2010}>\frac{1}{2011}\Rightarrow\frac{2010}{2011}>\frac{2009}{2010}\)
b)\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)
\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)
Vì \(\frac{1}{81^{100}}<\frac{1}{64^{100}}\Rightarrow\)\(\frac{1}{3^{400}}<\frac{1}{4^{300}}\)
c)\(\frac{2008}{2008.2009}=\frac{1}{2009};\frac{2009}{2009.2010}=\frac{1}{2010}\)
vì \(\frac{1}{2009}>\frac{1}{2010}\Rightarrow\frac{2008}{2008.2009}>\frac{2009}{2009.2010}\)
d)tính tổng hai vế rồi so sánh
\(\frac{2010}{1}+\frac{2009}{2}+...+\frac{1}{2010}\)
\(=\left(1+1+...+1\right)+\frac{2009}{2}+...+\frac{1}{2010}\)
\(=\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1\)
\(=\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}\)
\(=2011.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)\)
Ta thấy tử số gấp 2011 lần mẫu số nên phép tính này=2011