Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: AN=NM
=>N là trung điểm của AM
=>\(MN=\frac12\times MA\)
=>\(S_{MBN}=\frac12\times S_{ABM}\)
Ta có: BM+MC=BC
=>\(BC=MC+\frac12\times MC=\frac32\times MC\)
=>\(\frac{BM}{BC}=\frac13\)
=>\(S_{ABM}=\frac13\times S_{ABC}\)
=>\(S_{MBN}=\frac16\times S_{ABC}\)
b: Ta có: \(AN=\frac12\times AM\)
=>\(S_{BNA}=\frac12\times S_{BMA}=\frac12\times\frac13\times S_{ABC}=\frac16\times S_{ABC}\)
Ta có: \(MC=\frac23\times BC\)
=>\(S_{CMA}=\frac23\times S_{CAB}\)
Ta có: \(MN=\frac12\times MA\)
=>\(S_{CMN}=\frac12\times S_{CMA}=\frac12\times\frac23\times S_{ABC}=\frac13\times S_{ABC}\)
\(S_{BNC}=S_{BNM}+S_{MNC}\)
\(=\frac16\times S_{ABC}+\frac13\times S_{ABC}=\frac12\times S_{ABC}\)
=>\(\frac{S_{BNC}}{S_{BNA}}=\frac12:\frac16=3\)
Ta có: I nằm giữa A và C
=>\(\frac{S_{BIA}}{S_{BIC}}=\frac{IA}{IC};\frac{S_{NIA}}{S_{NIC}}=\frac{IA}{IC}\)
=>\(\frac{S_{BIA}-S_{NIA}}{S_{BIC}-S_{NIC}}=\frac{IA}{IC}\)
=>\(\frac{IA}{IC}=\frac{S_{BNA}}{S_{BNC}}=\frac13\)
=>\(IC=3\times IA\)
\(A=1+\frac{1}{2}+...+\frac{1}{16}\)
= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)
> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
= \(1+2\times\frac{13}{12}\)
= \(1+\frac{13}{6}\)
= \(1+2+\frac{1}{6}\)
= \(3+\frac{1}{6}\)>\(3\)
=> \(A>3+\frac{1}{6}>3\)
=> \(A>3+\frac{1}{6}>B\)
=> \(A>B\)
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
A=2003x2004 - 1/2003x2004=2003x2004/2003x2004 - 1/2003x2004=1 - 1/2003x2004
B=2004x2005 - 1/2004x2005=2004x2005/2004x2005 - 1/2004x2005=1 - 1/2004x2005
Vì 1=1 và 1/2003x2004 > 1/2004x2005 nên 1-1/2003x2004 < 1-1/2004x2005
Vậy B < A
MÌnh ấn nhấm B > A