A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)

A=\(\dfrac{1}{1}-\dfrac{1}{39}\)

A=\(\dfrac{38}{39}\)

còn lại tự làm do mình có việc chút

Chưa học

\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)

2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))

2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)

A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)

Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)

Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )

Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)

Ta có :

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..............+\dfrac{1}{98.99.100}\)

\(S=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+................+\dfrac{2}{98.99.100}\right)\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)

\(S=\dfrac{1}{2}.\dfrac{4949}{9900}\)

\(S=\dfrac{4949}{19800}\)

~ Chúc bn học tốt ~

E=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

* Áp dụng công thức: \(\dfrac{k}{n.\left(n+k\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+k}\)

ta có : \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

E=\(\dfrac{1}{1.2}-\dfrac{1}{99.100}\)

E= ........(tính ra)

E=4949/9900

Câu hỏi của Dung Van - Toán lớp 6 | Học trực tuyến tìm kĩ trước khi hỏi nhé.

a) Ta có:

3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)

A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)

Lấy (1) - (2) ta được:

1-\(\dfrac{1}{3^{100}}\)

b) Ta xét:

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)

Ta có:

2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)

=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)

=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)

Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)

=\(\dfrac{370}{741}\)

Nếu bn cảm thấy mk đúng tick cho mk nhé!

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}\)

\(=\dfrac{65}{264}\)

Vậy...

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

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{20}\)

\(A=\dfrac{20}{20}-\dfrac{1}{20}\)

\(A=\dfrac{19}{20}\)

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

A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{18.19}\)-\(\dfrac{1}{19.20}\)

A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)

A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)

A = \(\dfrac{189}{380}\)

(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha )

\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt

A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\)

Theo công thức:

\(\dfrac{2m}{b.\left(b+m\right).\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+m.2\right)}\)Ta có:

2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{18.19.20}\)

2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)2A=\(\dfrac{1}{1.2}-\dfrac{1}{19.20}\)

2A=\(\dfrac{1}{2}-\dfrac{1}{19.20}\)

A=\(\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\)

A=\(\dfrac{1}{2}.\dfrac{19.20-2}{2.19.20}\)

A=\(\dfrac{19.20-2}{2.2.19.20}\) < \(\dfrac{19.20}{2.2.19.20}\) = \(\dfrac{1}{4}\)

\(\Rightarrow\) A<\(\dfrac{1}{4}\)