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Ta có:
n = \(\frac{2003+2004}{2004+2005}\)
\(=>\) n = \(\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)
Vì \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)
\(\frac{2004}{2005}>\frac{2004}{2004+2005}\)
\(=>\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)
\(=>\)m > n
Chúc bạn học tốt :)
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 \(\frac{1}{2003\times2004}>\frac{1}{2004\times2005}\Rightarrow A< B\)
\(\frac{2005\cdot2004-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot\left(2003+1\right)-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2005-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2004}{2003\cdot2005+2004}\)
\(=1\)
2005 x 2004 - 1 / 2003 × 2005 + 2004
= 2005 × (2003 + 1) - 1 / 2003 × 2005 + 2004
= 2005 × 2003 + (2005 - 1) / 2003 × 2005 + 2004
= 2005 × 2003 + 2004 / 2003 × 2005 + 2004
= 1
\(\frac{2005\times2004-1}{2003\times2005+2004}=\frac{2005\times2003+2005-1}{2003\times2005+2004}=\frac{2005\times2003+2004}{2003\times2005+2004}=1\)
\(\frac{1999}{2004}+\frac{2001}{2005}+\frac{5}{2004}+\frac{4}{2005}\)
=\(\left(\frac{1999}{2004}+\frac{5}{2004}\right)+\left(\frac{2001}{2005}+\frac{4}{2005}\right)\)
=\(1+1\)
=2
\(\frac{1999}{2004}+\frac{2001}{2005}+\frac{5}{2004}+\frac{4}{2005}\) = \(\frac{1999}{2004}+\frac{5}{2004}+\frac{2001}{2005}+\frac{4}{2005}\)
= \(1+1\)
= \(2\)
Click mik nha !!!!!!!!!!!!!!!!!!!
\(B=\frac{2003+2004}{2004+2005}=\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)
Ta có: \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)
\(\frac{2004}{2005}>\frac{2004}{2004+2005}\)
\(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2005}\)
\(A>B\)
Vậy A>B
đáp án là A>B
chúc bn hok tốt!