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\(A=\frac{2009.2010-2}{2008+2008.2010}=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
\(=\frac{2008.2010+2010-2}{2008+2008.2010}=\frac{2008.2010+2008}{2008+2008.2010}=1\)
\(B=\frac{-2009.20102010}{20092009.2010}=\frac{-2009.2010.10001}{2009.10001.2010}=-1\)
Vậy A + B = 1 + (-1) = 0
\(A=\frac{2009.2010-2}{2008+2008.2010}=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
\(=\frac{2008.2010+2010-2}{2008+2008.2010}=\frac{2008.2010+2008}{2008+2008.2010}=1\)
\(B=\frac{-2009.20102010}{20092009.2010}=\frac{-2009.2010.10001}{2009.10001.2010}\)
Vậy A + B = 1 + (-1) = 0
Ta có : \(A=\frac{2009.2010-2}{2008+2008.2010}>0\)
\(B=\frac{-2009.20102010}{20092009.2010}< 0\)
Nên A > B
Ta có :
\(A=\frac{2009.2010-2}{2008+2008.2010}\)
\(A=\frac{\left(2008+1\right).2010-2}{2008.\left(1+2010\right)}=\frac{2008.2010+2008}{2008.2011}\)
\(A=\frac{2008.\left(1+2010\right)}{2008.2011}=\frac{2008.2011}{2008.2011}=1\)
\(B=\frac{-2009.20102010}{20092009.2010}=\frac{\left(-2009\right).2010.10001}{2009.10001.2010}=\frac{-2009}{2009}=-1\)
Vậy \(A+B=1+\left(-1\right)=0\)
\(A=\frac{2009\cdot2010-2}{2008+2008\cdot2010}=\frac{\left(2008+1\right)\cdot2010-2}{2008+2008\cdot2010}=\frac{\left(2008\cdot2010\right)+\left(2010-2\right)}{2008+2008\cdot2010}=\frac{2008\cdot2010+2008}{2008+2008\cdot2010}=1\)\(B=\frac{-2009\cdot20102010}{20092009\cdot2010}=\frac{-2009\cdot2010\cdot10001}{2009\cdot10001\cdot2010}=-1\)
=>A+B=1+(-1)=0
Vậy A+B=0
\(A=\frac{2009.2010-2}{2008+2008.2010}=\frac{2008.2010+2010-2}{2008+2008.2010}=\frac{2008.2010+2008}{2008+2008.2010}=1\)
\(B=\frac{-2009.20102010}{20092009.2010}=\frac{-2009.2010.10001}{2009.10001.2010}=-1\)
Vậy \(a+b=1+\left(-1\right)=0\)
A=\(\frac{2009.2010-2}{2008+2008.2010}=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
\(=\frac{2008.2010+2010-2}{2008+2008.2010}\)
\(=\frac{2008.2010+2008}{2008+2008.2010}=1\)
B=\(\frac{-2009.20102010}{20092009.2010}=\frac{-2009.2010.10001}{2009.10001.2010}=-1\)
Vậy A+|B= 1+(-1)=0
A = \(\frac{2009.2010-2}{2008+2008.2010}=\frac{2009.2010-2}{2008.\left(2010+1\right)}=\frac{2009.2010-2}{2008.2011}=\frac{2008.2010+2010-2}{2008.2011}=\frac{2008.2011}{2008.2011}=1\)
B = \(\frac{-2009.20102010}{20092009.2010}=\frac{-2009.10001.2010}{2009.10001.2010}=-1\)
1 > -1 => A > B
Ta có:
\(A=\frac{2009.2010-2}{2008+2008.2010}\)
\(A=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
\(A=\frac{2008.2010+2010-2}{2008+2008.2010}\)
\(A=\frac{2008.2010+2008}{2008+2008.2010}\)
\(A=1\)
\(B=\frac{-2009.20102010}{20092009.2010}\)
\(B=\frac{-2009.2010.10001}{2009.10001.2010}\)
\(B=-1\)
Vì \(1>-1\Rightarrow A>B\)
Vậy \(A>B\)
A=\(\frac{2009.2010-2}{2008+2008.2010}\)=\(\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
A=\(\frac{2008.2010+2010-2}{2008+2008.2010}\)=\(\frac{2008.2010+2008}{2008+2008.2010}\)=1
Ta có;
B=\(\frac{-2009.20102010}{20092009.2010}\)=\(\frac{-2009.2010.10001}{2009.10001.2010}\)=-1
Vậy A+B=1+(-1)=0
robert lewandoski copy bài vì bài đâu có hỏi A+B
ai đồng ý thì l-i-k-e