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1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
\(A=1+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}\)
\(A=1+2^2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+..+\frac{1}{2009^2}\right)\)
Ta có: \(\frac{1}{3^2}< \frac{1}{1.3};\frac{1}{5^2}< \frac{1}{3.5};\frac{1}{7^2}< \frac{1}{5.7};...;\frac{1}{2009^2}< \frac{1}{2007.2009}\)
\(\Rightarrow A< 1+4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{2007.2009}\right)\)
\(=1+4\cdot\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=1+2\left(1-\frac{1}{2009}\right)=3-\frac{2}{2009}< 3\)
\(\Rightarrow A< 3\)
a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)
\(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B Vay A<B
b,lam tuong tu nhu y a
2/1*2*3+2/3*4*5+...+2/2009*2010*2011
A=2/2*(1/1-1/2-1/3+1/2-1/3-1/4+1/4-1/5-1/6+...+1/2009-1/2010-1/2011
A=1*(1-1/2011)
A=1*2010/2011=2010/2011
suy ra: 2010/2011<1
suy ra 1/2 của 1 lớn hơn 2010/2011
VẬY A NHỎ HƠN 1/2
VẬY
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
Ta có:
\(\frac{2^2}{3^2}>\frac{2^2}{2009^2}\)
\(\frac{2^2}{5^2}>\frac{2^2}{2009^2}\)
\(\frac{2^2}{7^2}>\frac{2^2}{2009^2}\)
.........
\(\frac{2^2}{2009^2}=\frac{2^2}{2009^2}\)
\(\Rightarrow\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}>\frac{2^2}{2009^2}+\frac{2^2}{2009^2}+\frac{2^2}{2009^2}+...+\frac{2^2}{2009^2}=\frac{2^2.1004}{2009^2}=\frac{4016}{2009^2}\)(1004 phân số \(\frac{2^2}{2009^2}\)) . Mà:
\(\frac{4016}{2009^2}< 3\)
=> A < 3