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A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 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 \(A>B\)
\(B=\frac{18}{37}-\frac{8}{2017}+\frac{19}{37}-1\frac{2009}{2017}+\frac{2017}{2018}\)
\(B=\left(\frac{18}{37}+\frac{19}{37}\right)-\left(\frac{8}{2017}+1\frac{2009}{2017}\right)+\frac{2017}{2018}\)
\(B=1-\left(\frac{8}{2017}+\frac{4026}{2017}\right)+\frac{2017}{2018}\)
\(B=1-2+\frac{2017}{2018}\)
\(B=-1+\frac{2017}{2018}=\frac{-2018}{2018}+\frac{2017}{2018}\)
\(B=\frac{-1}{2018}\)
CHÚC BN HỌC TỐT!!!!!!
mk nha!!
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
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
do \(2009^{2009}-2< 2009^{2010}-2\Rightarrow B< 1\)
theo bài ra ta có:
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(1+2009^{2009}\right)}{2009\left(1+2009^{2010}\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=A\Rightarrow B< A\)
chúc bạn học tốt!!!
Ta có B có tử và mẫu bằng nhau=> B = 1
Vi 20092009<20092010=>20092009+1<20092010+1
Vậy A có từ< mẫu=>A<1
=>A<B
VẬy A<B
Kết bạn với mình nhé
Ta có :
\(\frac{1}{2009}A=\frac{2009^{2017}+1}{2009^{2017}+2009}=\frac{2009^{2017}+2009}{2009^{2017}+2009}-\frac{2008}{2009^{2017}+2009}=1-\frac{2008}{2009^{2017}+2009}< 1\)
\(\frac{1}{2009}B=\frac{2009^{2018}-2}{2009^{2018}-4018}=\frac{2009^{2018}-4018}{2009^{2018}-4018}+\frac{4016}{2009^{2018}-4018}=1+\frac{4016}{2009^{2018}-4018}>1\)
\(\Rightarrow\)\(A< 1< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
Vi 2009^2017 + 1 / 2009^2016 + 1 > 1
nen 2009^2018 + 1 / 2009^2016 + 1 < 2009^2018 + 1 - 3 / 2009^2016 + 1 - 3 = 2009^2018 - 2 / 2009^2017 - 2
Vay ...
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