Bài 6: So sánh phân số

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=\dfrac{1}{1}-\dfrac{1}{14}=\dfrac{13}{14}\)

Ta thấy: \(\dfrac{13}{14}< 1\)

Vậy A<1

Đặt \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{14}{14}-\dfrac{1}{14}\)

\(=\dfrac{13}{14}< 1\)

Vậy A < 1

Ta có :B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

Vì \(1-\dfrac{1}{14}< 1\)

Suy ra : \(B< 1\)

Ta thấy: \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(...\)

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{49}{100}< \dfrac{50}{100}< \dfrac{1}{2}\) (ĐPCM)

\(\dfrac{-49}{78}\)<\(\dfrac{0}{78}\)=0=\(\dfrac{0}{-95}\)<\(\dfrac{64}{-95}\)

hông chắc nha :)

b/ sao mà so sánh cùng một lúc bốn phân số được bạn :(

bạn nào giải được giúp mình với

Help me pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!!!!!!!!!!!

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ \Rightarrow2A=2+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\\ \Rightarrow2A-A=2-\dfrac{1}{50^2}\\ A=\dfrac{5000}{50^2}-\dfrac{1}{50^2}=\dfrac{5000-1}{50^2}=\dfrac{4999}{2500}\)

Giả sử \(B=\dfrac{173}{100}=\dfrac{4325}{2500}\), mà \(\dfrac{4999}{2500}>\dfrac{4325}{2500}\)

\(\Rightarrow A>B\)

Có : \(2009+2010>\dfrac{2009}{2010}\) ; \(2011+2012>\dfrac{2011}{2012}\)

\(\dfrac{2011}{2010}>1\) ; \(\dfrac{2010}{2011}< 1\) \(\Rightarrow\dfrac{2011}{2010}>\dfrac{2010}{2011}\)

Ta có : \(2009+2010+\dfrac{2011}{2010}+2011+2012>\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2011}{2012}\)

\(\Leftrightarrow B>A\)

Hay \(A< B\)

\(A=\dfrac{2004^{2005}+1}{2004^{2005}-2004}>1>\dfrac{2004^{2005}}{2004^{2005}+2004}=B\)

Vậy A > B

Ta có :

\(\dfrac{2004^{2005}+1}{2004^{2005}-2004}>1>\dfrac{2004^{2005}}{2004^{2005}+2004}\)

\(\Rightarrow\) \(A>1>B\)

\(\Rightarrow\) \(A>B\)

\(\dfrac{2011\cdot2012-1}{2011\cdot2012}=\dfrac{2011\cdot2012}{2011\cdot2012}-\dfrac{1}{2011\cdot2012}=1-\dfrac{1}{2011\cdot2012}\)

\(\dfrac{2012\cdot2013-1}{2012\cdot2013}=\dfrac{2012\cdot2013}{2012\cdot2013}-\dfrac{1}{2012\cdot2013}=1-\dfrac{1}{2012\cdot2013}\)

Vì \(\dfrac{1}{2011\cdot2012}>\dfrac{1}{2012\cdot2013}\Rightarrow1-\dfrac{1}{2011\cdot2012}>1-\dfrac{1}{2012\cdot2013}\)

Vậy \(\dfrac{2011\cdot2012-1}{2011\cdot2012}< \dfrac{2012\cdot2013-1}{2012\cdot2013}\)

Ta có A = \(\dfrac{1919.161616}{323232.3838}=\dfrac{1919.161616}{2.1919.2.161616}=\dfrac{1}{2.2}=\dfrac{1}{4}\)

Vì\(\dfrac{1}{4}=\dfrac{25}{100}\) , mà \(\dfrac{25}{100}>\dfrac{25}{102}=>A>B\)

Ta có : \(A=\dfrac{1919\cdot161616}{323232\cdot3838}=\dfrac{1\cdot1}{2\cdot2}=\dfrac{1}{4}\)

Giả sử \(A=\dfrac{1}{4}=\dfrac{25}{100}\), mà \(\dfrac{25}{100}>\dfrac{25}{102}\)

\(\Rightarrow\) \(\dfrac{1}{4}>\dfrac{25}{102}\)

\(\Rightarrow\) \(A>B.\)

Ta có:A-1=\(\dfrac{10^8+2}{10^8-1}-1=\dfrac{10^8+2-10^8+1}{10^8-1}=\dfrac{3}{10^8-1}\)

B-1=\(\dfrac{10^8}{10^8-3}-1=\dfrac{10^8-10^8+3}{10^8-3}=\dfrac{3}{10^8-3}\)

Do \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\)

=>A-1>B-1

<=>A>B

Vậy...

mik cũng đg cần

\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)

\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)

Vì \(\dfrac{3}{10^8-1}< \dfrac{3}{10^8-3}\)

Nên \(1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\)

Vậy A < B.

Sai đề rồi \(\dfrac{1}{13}\) chứ đâu phải \(\dfrac{1}{3}\)

Ta có: \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

\(< \dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{61}+\dfrac{1}{61}+\dfrac{1}{61}\right)\)

\(=\dfrac{1}{5}+\dfrac{1}{13}.3+\dfrac{1}{61}.3\)

\(=\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)

\(< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}\)

\(=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(=\dfrac{10}{20}\)\(=\dfrac{1}{2}\)

Vậy S\(< \dfrac{1}{2}\) (đpcm)