Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Rightarrow B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Suy ra : \(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2009}+1}{2009^{2010}+1}\) hay \(B< A\)
Vậy \(A>B\)
Do 2009\(^{2010}\)-2 < 2009\(^{2011}\)-2 \(\Rightarrow\)B<1
Theo đề bài ta có:
B= \(\frac{2009^{2010}-2}{2009^{2011}-2}\)< \(\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)= \(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)= \(\frac{2009.\left(1+2009^{2009}\right)}{2009.\left(1+2009^{2010}\right)}\)= \(\frac{2009^{2009}+1}{2009^{2010}+1}\)= A \(\Rightarrow\)B<A
2009A=2009^2010+2009/2009^2010+1 2009B=2009^2011-4018/2009^2011-2
2009A=1 + 2009/2009^2010+1 B=1 - 4016/2009^2011-2
mình viết tách ra cho khỏi nhầm
vì A>1 và B<1
nên A>B
VẬY A>B AND kết bạn nha
A=2009^2009+1/2009^2010+1 B=2009^2010-2/2009^2011-2
A=(2009^2009+1).10/2009^2010+1 B=(2009^2010-2).10/2009^2011-2
A=2009^2010+10/2009^2010+1 B= 2009^2011-20/2009^2010-2
A=(2009^2010+1)+9/2009^2010+1 B=(2009^2011-2)-18/2009^2010-2
A=1 + 9/2009^2010+1 B=1+(-18/2009^2010-2)
Vì 9/2009^2010+1 > (-18/2009^2010-2)
=>1 + 9/2009^2010+1>1+(-18/2009^2010-2)
Hay 2009^2009+1/2009^2010+1 > 2009^2010-2/2009^2011-2
Vậy A>B
B = 2009^2010 - 2 / 2009^2011 - 2 < 2009^2010 - 2 + 2011 /2009^2011 - 2 + 2011
= 2009^2010 + 2009 / 2009^2011 + 2009
= 2009 ( 2009^2009 + 1) / 2009(2009^2010 + 1)
= 2009^2009 + 1 / 2009^2010 + 1 = A
=> B < A
B=20092010-2/20092011-2<20092010-2+2011/20092011-2+2011=20092010+2009/20092011+2009 =2009.(20092009+1)/2009.(20092010+1)=20092009+1/20092010+1
Suy ra A>B
=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
A=\(\dfrac{2009^{2010}+1}{2009^{2009}+1}\)
2009A=\(\dfrac{(2009^{2010}+1)+0}{2009^{2010}+1}\)
= 1+\(\dfrac{0}{2009^{2010}+1}\)= 1+0 =1
B=\(\dfrac{2009^{2011}-2}{2009^{2010}-2}\)
2009B=\(\dfrac{2009^{2011}-1}{2009^{2011}-2009}\)
=\(\dfrac{(2009^{2011}-1)-0}{2009^{2011}-2009}\)
= \(1-\dfrac{0}{2009^{2011}-2009}\)
=1-0= 1
Vì 1=1\(\Rightarrow A=B\)
Ta có : A = 2009^2010+1/2009^2009+1
Suy ra: 1/2009 A = 1 - 2008/2009^2010+2009 (1)
Lại có:B = 2009^2011 - 2 / 2009^2010 - 2
Suy ra : 1/2009 B = 1 + 4016/2009^2011-4018 (2)
Vì 1 - 2008/2009^2010+2009 < 1 + 4016/2009^2011-4018 (3)
Từ (1);(2) và (3) suy ra : A<B