Ghi lộn đề thiếu thì phải. Hình như thiếu phân số 1/2011

a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1

  = 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011

  = 2011(1/2+1/3+1/4+...+1/2011)

Ta có: B= 1/2+1/3+1/4+...+1/2011

suy ra A/B= 2011

=1/2010

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

Cho cách giải lun

a)

\(S=1-2+3-4+...+2009-2010\)

\(S=\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)

\(S=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

Có:

\(\dfrac{\left(2010-1\right):1+1}{2}=1005\) số (-1)

\(\Rightarrow S=1005.\left(-1\right)=-1005\)

b)

\(P=0-2+4-6+...+2010-2012\)

\(P=\left(0-2\right)+\left(4-6\right)+...+\left(2010-2012\right)\)

\(P=\left(-2\right)+\left(-2\right)+...+\left(-2\right)\)

Có:

\(\dfrac{\left(2010-0\right):2+1}{2}=503\) số (-2)

\(\Rightarrow P=503.\left(-2\right)=-1006\)

a)

Có:

số (-1)

b)

Có:

số (-2)

Ta có: A=\(\frac{1}{2011}+\frac{2}{2010}+\frac{3}{2009}+...+\frac{2009}{3}+\frac{2010}{2}+\frac{2011}{1}\)

=> A=\(\frac{2012-2011}{2011}+\frac{2012-2010}{2010}+...+\frac{2012-2}{2}+\frac{2012-1}{1}\)

=>A=\(\frac{2012}{2011}-1+\frac{2012}{2010}-1+...+\frac{2012}{2}-1+2012-1\)

=>A=\(2012\cdot\left(\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{2}\right)+1\)

=> A= \(2012\cdot\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{2}\right)\)

ko biết có đúng hay ko nựa sai thì bỏ qua nha ^^

dung r bn oi

con co cau p=1/2+1/3+...+1/2011+1/2012

1.

\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

cứ làm như vậy ta được :

\(=1+1=2\)

2. Ta có :

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

vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)

\(B=\frac{2009^{2010}-2}{2009^{2011}-2}< 1\)

\(\Rightarrow B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)

Suy ra : \(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2009}+1}{2009^{2010}+1}\) hay \(B< A\)

Vậy \(A>B\)

Bạn giải cũng được đấy alibaba nguyễn, nhưng theo mình thì làm cách này dễ hiểu hơn!

Ta có: \(C=\frac{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)

Đặt \(A=\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}\)

\(A=\frac{2010}{1}+1+\frac{2009}{1}+1+\frac{2008}{1}+1+...+\frac{1}{2010}+1-2010\)

\(=\frac{2011}{1}+\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}-\frac{2011.2010}{2011}\)

\(=2011\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-1\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\)

Ta có: \(C=\frac{A}{B}=2011\)(lấy A-B)

Ta có :

\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=\dfrac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)

\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=\dfrac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)

Vì \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)

~ Học tốt ~