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A=2003x2004-1/2003x2004
B=2004x2005-1/2004x2005
A= 1-2003x2004-1/2003x2004=1/2003x2004
B=1-2004x2005-1/2004x2005=1/2004x2005
Vì 1/2003x2004<1/2004x2005 => A>B.
K nhé
Đáp án là B lớn hơn A nha
NHỚ K CHO MIK NHA MY FRIEND :>
A=192x198
A=192x (197+1)
A=192x197+192
B=193x197
B=197x(192+1)
B=197x192+197
Có A=192x197+192 < B=197x192+197
nên A<B
K nha
A=(193-1)*198=193*198-198
B=193*(198-1)=193*198-193
=>A<B
\(A=1+\frac{1}{2}+...+\frac{1}{16}\)
= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)
> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
= \(1+2\times\frac{13}{12}\)
= \(1+\frac{13}{6}\)
= \(1+2+\frac{1}{6}\)
= \(3+\frac{1}{6}\)>\(3\)
=> \(A>3+\frac{1}{6}>3\)
=> \(A>3+\frac{1}{6}>B\)
=> \(A>B\)
a)\(\frac{14}{15}\) < \(\frac{15}{21}\)
b)\(\frac{101}{200}\) < \(\frac{200}{404}\)
c)\(\frac{1995}{2011}\) >\(\frac{1993}{2012}\)
#)Giải :
Ta có :
\(A=\frac{2003\times2004-1}{2003\times2004}=\frac{2003\times2004}{2003\times2004}-\frac{1}{2003\times2004}=1-\frac{1}{2003\times2004}\)
\(B=\frac{2004\times2005-1}{2004\times2005}=\frac{2004\times2005}{2004\times2005}-\frac{1}{2004\times2005}=1-\frac{1}{2004\times2005}\)
Vì \(\frac{1}{2003\times2004}>\frac{1}{2004\times2005}\)
\(\Rightarrow A>B\)
+) \(A=\frac{2003\times2004-1}{2003\times2004}\)
\(=\frac{2003\times2004}{2003\times2004}-\frac{1}{2003\times2004}\)
\(=1-\frac{1}{2003\times2004}\)
+) \(B=\frac{2004\times2005-1}{2004\times2005}\)
\(=\frac{2004\times2005}{2004\times2005}-\frac{1}{2004\times2005}\)
\(=1-\frac{1}{2004\times2005}\)
+) Vì 2004 x 2005 > 2003 x 2004
=> \(\frac{1}{2004\times2005}< \frac{1}{2003\times2004}\)
=> \(1-\frac{1}{2004\times2005}>1-\frac{1}{2003\times2004}\)
Vậy B > A