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\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}<\frac{2011}{2012}+\frac{2012}{2013}=A\)
vậy A>B
\(A=\frac{2011}{2012}+\frac{2012}{2013}\) \(và\) \(B=\frac{2011+2012}{2012+2013}\)
\(Ta\) \(có\) \(:\) \(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}\)
\(B=\frac{2011}{4025}+\frac{2012}{4025}\)
\(Vì\) \(\frac{2011}{2012}>\frac{2011}{4025}và\frac{2012}{2013}>\frac{2012}{4025}\)
\(Nên\) \(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{4025}+\frac{2012}{4025}\)
\(Vậy\) \(A=\frac{2011}{2012}+\frac{2012}{2013}>B=\frac{2011+2012}{2012+2013}\)
98 <1
99
98.99+1 Vì 98.99+1 >98.99 nên 98.99+1 >1
98.99 98.99
Suy ra: 98 < 98.99+1
99 98.99
A= \(\frac{98}{99}\)< \(1\)
B= \(\frac{98.99+1}{98.99}\)=\(\frac{98.99}{98.99}+\frac{1}{98.99}\)=\(1+\frac{1}{98.99}\)> 1
=> A<1<B => A<B
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Ta có :
\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)
\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)
\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)
\(\Leftrightarrow B< \frac{1}{4}\)
B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)
Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)
\(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)
...
Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)
Vậy: B<\(\frac{1}{4}\)
ta có A=\(\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}\)(1)
B=\(\frac{2011}{2012}+\frac{2012}{2013}\left(2\right)\)
so sánh 1 và 2 ta có A<B
B=2011+2012/2012+2013
=2011/2012+2013 +2012/2012+2013<2011/2012 +2012/2013=a
vậy........................
B=3^10.11+3^10.5/3^9.2^4
= 3^10( 11+5)/3^9.16
= 3^10.16/3^9.16
= 3^10/3^9
= 3
Vậy B = 3 (1)
C = 2^10.13+2^10.65/2^8.104
= 2^10(13+65)/2^8.2^2.26
= 2^10.78/2^10.26
= 78/26
= 3
Vậy C = 3 (2)
Từ (1) v (2) suy ra B=C
Câu 1 :
Ta có : \(A=\frac{10^{100}+1}{10^{101}+1}\)
\(\Rightarrow10A=\frac{10^{101}+10}{10^{101}+1}=\frac{10^{101}+1+9}{10^{101}+1}=1+\frac{9}{10^{101}+1}\)
Ta có : \(B=\frac{10^{101}+1}{10^{102}+1}\)
\(10B=\frac{10^{102}+10}{10^{102}+1}=\frac{10^{102}+1+9}{10^{102}+1}=1+\frac{9}{10^{102}+1}\)
Vì 10101+1<10102+1
\(\Rightarrow\frac{9}{10^{101}+1}>\frac{9}{10^{102}+1}\)
\(\Rightarrow1+\frac{9}{10^{101}+1}>1+\frac{9}{10^{102}+1}\)
\(\Rightarrow\)10A>10B
\(\Rightarrow\)A>B
Vậy A>B.
Câu 2 :
Ta có : \(E=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì 2001<2001+2002 và 2002<2001+2002
\(\Rightarrow\hept{\begin{cases}\frac{2000}{2001}>\frac{2000}{2001+2002}\\\frac{2001}{2002}>\frac{2001}{2001+2002}\end{cases}}\)
\(\Rightarrow C>E\)
Vậy C>E.
phần bù đến 1 của 21/31 là: 1-21/31=10/31
phần bù đến 1 của 217/317 là:1-217/317=100/317
ta có 10/31=100/310
vì 100/310>100/317 nên 10/31>100/317 => 21/31<217/317 => A<B