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10 tháng 4 2016

Tổng quát: \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\) (với mọi số tự nhiên n khác 0)

Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}<\frac{1}{2}\) (vì \(\frac{1}{100}>0\) )

=>đpcm

 

15 tháng 4 2016
\(\frac{9.25-63}{9.10+153}\)=\(\frac{9.25-9.7}{9.10+9.17}\)=\(\frac{9.\left(25-7\right)}{9.\left(10+17\right)}\)=\(\frac{9.18}{9.27}\)=\(\frac{1.2}{1.3}\)=\(\frac{2}{3}\)

   1/1.2 + 1/2.3 + 1/3.4 + .......................+ 1/99.100    

= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 +..................+ 1/99 - 1/100 

= 1 - 1/100 

= 99/100 

30 tháng 4 2016

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

Ma 99/100 < 1.

=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1 (dccm)

13 tháng 4 2018

\(b)\) Đặt \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) ta có : 

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}-0=\frac{1}{2}\)

\(\Rightarrow\)\(A< \frac{1}{2}\) ( đpcm ) 

Vậy \(A< \frac{1}{2}\)

Chúc bạn học tốt ~ 

13 tháng 4 2018

\(a)\frac{9.25-63}{3.30+153}\)

\(=\frac{9.25-9.7}{3.30+3.51}\)

\(=\frac{9.\left(25-7\right)}{3.\left(30+51\right)}\)

\(=\frac{9.18}{3.81}\)

\(=\frac{1.6}{1.9}\)

\(=\frac{6}{9}\)

\(=\frac{2}{3}\)

b )    \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(Đpcm\right)\)

Chúc bạn học tốt !!! 

22 tháng 4 2018

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{59\cdot60}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{69}-\frac{1}{60}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

11 tháng 4 2019

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

\(=1-\frac{1}{100}< 1\)

11 tháng 4 2019

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

22 tháng 7 2020

Bài 15 :

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)

b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)

\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)

Tới đây là so sánh đi nhé

22 tháng 7 2020

Cái này mình làm hôm qua rồi mà '-'

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A< 1\)

b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)

\(2A-A=A\)

\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)

\(=1-\frac{1}{2^{1000}}\)

\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)

2 tháng 4 2016

S = 1/2+1/2.3+1/3.4 +... +1/9/10

S =1/2+1/2-1/3+1/3+1/4+...+1/9-1/10

S =1-10

S =9/10

Do 9/10<1

=>S<1

2 tháng 4 2016

S=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

  =1-(1/2-1/2)-(1/3-1/3)-(1/4-1/4)-...-(1/9-1/9)-1/10

  =1-1/10<1

Vậy S<1