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20 tháng 3 2016

$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

24 tháng 3 2017

a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760

26 tháng 9 2017

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)

Cái B TT nhé

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)

D TT

E mk thấy nó ss ớ

26 tháng 9 2017

ai thế

10 tháng 1 2020

a=bao nhiêu bạn

10 tháng 1 2020

xin lỗi vì thếu đề 

a bài 1 là bằng 3/11 nha

20 tháng 7 2023

a/

\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)

b/

\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)

\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)

c/

\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)

đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)

Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)

\(=3-\frac{3}{20}=\frac{57}{20}\)

\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)

5 tháng 7 2017

Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A

\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)

\(A=\frac{1}{2}.\frac{19}{20}\)

\(A=\frac{19}{40}\)

\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)

\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)

\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)

\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(B=\left[3.\left(\frac{19}{20}\right)\right]\)

\(B=\frac{57}{20}\)

Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)

\(=-\frac{95}{40}=-\frac{19}{8}\)

Nếu đúng thì k nha

2 tháng 4 2016

khó à nha