\(A=\frac{2009.2010-2}{2008+2008.2010}=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)

     \(=\frac{2008.2010+2010-2}{2008+2008.2010}=\frac{2008.2010+2008}{2008+2008.2010}=1\)

\(B=\frac{-2009.20102010}{20092009.2010}=\frac{-2009.2010.10001}{2009.10001.2010}=-1\)

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

kiểu này cuối tuần việt đầu bảng điểm chắc lun

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 \(A+B=1+\left(-1\right)=0\)

Vì 2009*2010>2.

=>2009*2010-2>0.

2008*2008*2010>0(hiển nhiên).

=>A>0.

Ta dễ thấy:

Tử của B là số âm còn mẫu là số dương.

=>B<0.

=>A>0>B.

Vậy A>B.

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 \(A>B\)

Ta có -2009.20102010 bé hơn 0 =>B bé hơn 0 (1)

Lại có:2009.2010-2lonws hơn 0=> A lớn hơn 0 (2)

Từ (1) và (2) => A>B

\(A=\frac{2009\cdot2010-2}{2008+2008\cdot2010}=\frac{\left(2008+1\right)\cdot2010-2}{2008+2008\cdot2010}=\frac{\left(2008\cdot2010\right)+\left(2010-2\right)}{2008+2008\cdot2010}=\frac{2008\cdot2010+2008}{2008+2008\cdot2010}=1\)\(B=\frac{-2009\cdot20102010}{20092009\cdot2010}=\frac{-2009\cdot2010\cdot10001}{2009\cdot10001\cdot2010}=-1\)

=>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

vậy a+b=1+1=0

= 0 nha bạn 

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