Ta có :

+) \(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)

+) \(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)

ta thấy :

\(\frac{1}{2003.2004}>\frac{1}{2004.2005}\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)

\(\Rightarrow\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)

\(A=\dfrac{2003.2004-1}{2003.2004}=1-\dfrac{1}{2003.2004}\)

\(B=\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

So sánh: \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)

\(\Rightarrow-\dfrac{1}{2003.2004}< -\dfrac{1}{2004.2005}\\ \Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\\ Hay.A< B\)

a) A=\(\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2004}=1-\dfrac{1}{2003.2004}\)

B = \(\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)

\(\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)

Vậy A < B

b) \(\left(3X-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)

\(\left(3X-2^4\right).7^5=2.7^6.1\)

\(\left(3X-2^4\right).7^5=2.7^6\)

\(\left(3X-2^4\right).=2.7^6:7^5\)

\(3X-2^4=2.7\)

\(3X-16=14\)

\(3X=16+14=30\)

\(X=30:3=10\)

Vậy X = 10

1/ \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}=1-\dfrac{1}{2003.2004}\)

\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

Vì \(1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\Leftrightarrow A< B\)

2/ \(\left(3x-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)

\(\Leftrightarrow\left(3x-2^4\right).7^5=2.7^6.1\)

\(\Leftrightarrow3x-2^4=2.7^6:7^5\)

\(\Leftrightarrow3x-2^4=2.7\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\left(tm\right)\)

Vậy ..

a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)

a) Ta có: \(\dfrac{39}{-65}=\dfrac{-39}{65}=\dfrac{-39:13}{65:13}=\dfrac{-3}{5}\)

\(\dfrac{-3}{5}=\dfrac{-3}{5}\)

Do đó: \(\dfrac{-3}{5}=\dfrac{39}{-65}\)

b) Ta có: \(\dfrac{-9}{27}=\dfrac{-9:9}{27:9}=\dfrac{-1}{3}\)

\(\dfrac{-41}{123}=\dfrac{-41:41}{123:41}=\dfrac{-1}{3}\)

Do đó: \(\dfrac{-9}{27}=\dfrac{-41}{123}\)

c) Ta có: \(\dfrac{-3}{4}=\dfrac{-3\cdot5}{4\cdot5}=\dfrac{-15}{20}\)

\(\dfrac{4}{-5}=\dfrac{-4}{5}=\dfrac{-4\cdot4}{5\cdot4}=\dfrac{-16}{20}\)

mà \(\dfrac{-15}{20}>\dfrac{-16}{20}\)

nên \(\dfrac{-3}{4}>\dfrac{4}{-5}\)

d) Ta có: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot7}{3\cdot7}=\dfrac{-14}{21}\)

\(\dfrac{-5}{7}=\dfrac{-5\cdot3}{7\cdot3}=\dfrac{-15}{21}\)

mà \(\dfrac{-14}{21}>\dfrac{-15}{21}\)

nên \(\dfrac{2}{-3}>\dfrac{-5}{7}\)

b ) \(\frac{2013.2014-1}{2013.2014}=\frac{2013.2014}{2013.2014}-\frac{1}{2013.2014}=1-\frac{1}{2013.2014}\)

\(\frac{2014.2015-1}{2014.2015}=\frac{2014.2015}{2014.2015}-\frac{1}{2014.2015}=1-\frac{1}{2014.2015}\)

Vì \(2013.2014< 2014.2015\) nên \(\frac{1}{2013.2014}>\frac{1}{2014.2015}\)

Do đó \(1-\frac{1}{2013.2014}< 1-\frac{1}{2014.2015}\)

=> \(\frac{2013.2014-1}{2013.2014}< \frac{2014.2015-1}{2014.2015}\)

a) 1-\(\frac{67}{77}\)= \(\frac{10}{77}\);1-\(\frac{73}{83}\)= \(\frac{10}{83}\)Vậy \(\frac{10}{77}\)>\(\frac{10}{83}\)Nên \(\frac{67}{77}\)>\(\frac{73}{83}\)

Các câu khác cũng vậy mk cũng ko chắc câu đầu đâu

a: 2003.2004-1/2003.2004      <         2004.2005-1/2004.2005

b: 149/157 < 449/457

c: 1999.2000/1999.2000+1        <      2000.2001/2000.2001+1