Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010) 

=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010

=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0

=> -2010a + 2.2009b = 0

=> 2010a = 2.2009b

Đề sai

Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010)

=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010

=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0

=> -2010a + 2.2009b = 0

=> 2010a = 2.2009b Đề sai 

\(\Leftrightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\)

=>ab-2010a+2009b-2009x2010=ab+2010a-2009b-2009x2010

=>-4020a=-4018b

=>a/2009=b/2010

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)

\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)

mình cũng có bài giống như này nhưng chưa làm được

mình cũng thế

^{Ta có: 2008/2009 > 2008/2009+2010+2011}

       2009/2010> 2009/2010+2011

      2010/2011>2010>2010/2009+2010+2011

Suy ra: A>2008+2009+2010/2009+2010+2011

Vậy A >B

Tớ cũng có bài này nhưng chưa làm được

cau tra loi la 50 khong can biet lam the nao

\(A=\left(2010^{2009}+2009^{2009}\right)^{2010}\)

\(=\left(2010^{2009}+2009^{2009}\right)^{2009}\left(2010^{2009}+2009^{2009}\right)\)

\(>\left(2010^{2009}+2009^{2009}\right)^{2009}.2010^{2009}\)

\(=\left(2010.2010^{2009}+2010.2009^{2009}\right)^{2009}\)

\(>\left(2010.2010^{2009}+2009.2009^{2009}\right)^{2009}\)

\(=\left(2010^{2010}+2009^{2010}\right)^{2009}=B\)

Vậy \(A>B\)

Dạo này anh ít on lắm em có nhờ thì em kiếm kênh khác nhờ không thì phải đợi a on a mới làm được nhé

E có cách khác a ơi :)

\(\text{We Have:}\)

\(B=\left(2010^{2010}+2009^{2010}\right)^{2009}=\left(2010.2010^{2009}+2009.2009^{2009}\right)^{2009}\)

\(< \left(2010.2010^{2009}+2010.2009^{2009}\right)^{2009}\)

\(=\left[2010\left(2010^{2009}+2009^{2009}\right)\right]^{2009}=\left(2010^{2009}+2009^{2009}\right)^{2009}.2010^{2009}\)

\(< \left(2010^{2009}+2009^{2009}\right)^{2009}.\left(2010^{2009}+2009^{2009}\right)\)

\(=\left(2010^{2009}+2009^{2009}\right)^{2010}=A\Rightarrow A>B\)

\(\text{So: A is "lớn hơn" B =))}\)

Đặt \(a=2010^{2009};b=2009^{2009}\)\(\left(a,b>0\right)\)

\(A=\left(a+b\right)^{2010}=\left(a+b\right)^{2009}.\left(a+b\right)\)

\(B=\left(a.2010+b.2009\right)^{2009}=\left[a+2009\left(a+b\right)\right]^{2009}\)

Chia A và B cho \(\left(a+b\right)^{2009}:\)

\(A=a+b;B=\dfrac{\left[a+2009\left(a+b\right)\right]^{2009}}{\left(a+b\right)^{2009}}\)\(=\left(\dfrac{a}{a+b}+2009\right)^{2009}\)

Dễ thấy A<B.

\(B=\left(2010^{2009}.2010+2009^{2009}.2009\right)^{2009}\)

\(B< \left(2010^{2009}.2010+2009^{2009}.2010\right)^{2009}\)

\(B< \left(2010^{2009}+2009^{2009}\right)^{2009}.2010^{2009}\)

\(B< \left(2010^{2009}+2009^{2009}\right)^{2009}.\left(2010^{2009}+2009^{2009}\right)\)

\(B< \left(2010^{2009}+2009^{2009}\right)^{2010}\)

\(\Rightarrow B< A\)

A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)

Ta có: 

\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)

\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)

\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)

Từ 3 điều trên suy ra : A < B