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

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

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: \(\frac{23}{55}=1-\frac{32}{55};\frac{1978}{2010}=1-\frac{32}{2010}\)

Ta có: 55<2010

=>\(\frac{32}{55}>\frac{32}{2010}\)

=>\(-\frac{32}{55}<-\frac{32}{2010}\)

=>\(-\frac{32}{55}+1<-\frac{32}{2010}+1\)

=>\(\frac{23}{55}<\frac{1978}{2010}\)

b: \(\frac{2003\cdot2004-1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)

\(\frac{2004\cdot2005-1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)

Ta có: \(2003\cdot2004<2004\cdot2005\)

=>\(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}<\frac{-1}{2004\cdot2005}\)

=>\(\frac{-1}{2003\cdot2004}+1<\frac{-1}{2004\cdot2005}+1\)

=>\(\frac{2003\cdot2004-1}{2003\cdot2004}<\frac{2004\cdot2005-1}{2004\cdot2005}\)

c: \(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\)

=>\(\frac{a}{b}>\frac{a+m}{b+m}\)

d: a<b

=>a-b<0

\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}<0\)

=>\(\frac{a}{b}<\frac{a+m}{b+m}\)

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 :

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

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

Ta có:

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

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

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

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 ..

minh nghi la: 2003.2004-1 < 2004.2005-1

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