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\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)
\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)
\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{10}\)
\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)
\(\Leftrightarrow D< 1\left(đpcm\right)\)
Ta có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{1}{300}.200=\dfrac{200}{300}=\dfrac{2}{3}\)
\(\Rightarrow\) biểu thức trên lớn hơn \(\dfrac{2}{3}\).
a)
Ta thấy:
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
\(\dfrac{1}{12}< \dfrac{1}{10}\)
\(\dfrac{1}{13}< \dfrac{1}{10}\)
...
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 5\cdot\dfrac{1}{5}+8\cdot\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
b)
Ta thấy:
\(\dfrac{1}{101}>\dfrac{1}{300}\)
\(\dfrac{1}{102}>\dfrac{1}{300}\)
\(\dfrac{1}{103}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>200\cdot\dfrac{1}{300}=\dfrac{2}{3}\)
Vậy \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)
mọi người thật là nhẫn tâm
chẳng ai giúp mk
TRỜI ƠI!!! AI MS LÀ BN BÈ THỰC SỰ
Ko cs đứa mô trả lời chứ chi
Loại bn bè vs mấy ng chỉ là giả tạo thôi
Ta có :
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)
...
Do đó :
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
Vậy :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
hôm qua cô giảng cho mình bài này không cần tính đâu
Gọi tổng là A
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{17.18.19}\)
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{17.18.19}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{17.18}-\dfrac{1}{18.19}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{18.19}\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{18.19}\right)\)
A=\(\dfrac{1}{2}.\dfrac{18.19-2}{2.18.19}\) < \(\dfrac{1}{4}\)
A=\(\dfrac{18.19-2}{2.2.18.19}\) < \(\dfrac{18.19}{2.2.18.19}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
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 \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...