\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}+\dfrac{1}{18}+\dfra...">
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27 tháng 3 2017

A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)

A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)

A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)

A=\(\dfrac{7}{24}\)

B=\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)

B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)

B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)

B=\(1+\left(-1\right)+\left(-1\right)=-1\)

C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)

D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)

D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)

16 tháng 3 2018

a, Ta có :

\(\dfrac{1}{6}< \dfrac{1}{5}\)

\(\dfrac{1}{7}< \dfrac{1}{5}\)

.................

\(\dfrac{1}{9}< \dfrac{1}{5}\)

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

\(\dfrac{1}{11}< \dfrac{1}{10}\)

..................

\(\dfrac{1}{17}< \dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+......+\dfrac{1}{17}< \dfrac{1}{5}+\dfrac{1}{5}+....+\dfrac{1}{5}\)

\(\Leftrightarrow A< \dfrac{1}{5}.5+\dfrac{1}{10}.8\)

\(\Leftrightarrow A< 1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)

\(\Leftrightarrow A< 2\left(đpcm\right)\)

b/ Ta có :

\(\dfrac{1}{11}>\dfrac{1}{30}\)

\(\dfrac{1}{12}>\dfrac{1}{30}\)

...............

\(\dfrac{1}{29}>\dfrac{1}{30}\)

\(\dfrac{1}{30}=\dfrac{1}{30}\)

\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+........+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+.......+\dfrac{1}{30}\)

\(\Leftrightarrow B>\dfrac{1}{30}.20=\dfrac{2}{3}\)

\(\Leftrightarrow B>\dfrac{2}{3}\left(đpcm\right)\)

30 tháng 3 2017

Đặt :

\(A=\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...................+\dfrac{1}{17}\)

*Nhận xét :

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+.......................+\dfrac{1}{10}< \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...............+\dfrac{1}{5}\)

\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..............+\dfrac{1}{17}< \dfrac{1}{11}+\dfrac{1}{11}+.............+\dfrac{1}{11}\)

\(\Rightarrow A< \left(\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...........+\dfrac{1}{5}\right)+\left(\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+............+\dfrac{1}{11}\right)\)

\(\Rightarrow A< \left(\dfrac{1+1+1+................+1}{5}\right)+\left(\dfrac{1+1+1+..........+1}{11}\right)\)

\(\Rightarrow A< \dfrac{6}{5}+\dfrac{7}{11}\)

\(\Rightarrow A< \dfrac{110}{55}=2\)

\(\Rightarrow A< 2\)

Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+.................+\dfrac{1}{17}< 2\) \(\left(đpcm\right)\)

Chúc bn học tốt !!!!!!!!!

30 tháng 3 2017

thank you

5 tháng 5 2017

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 \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)

5 tháng 5 2017

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 \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)

11 tháng 4 2018

2,

\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{4}\)

7 tháng 4 2017

1/5<2;1/6<2;...;1/17<2

suy ra 1/5+1/6+1/7+...+1/17<2

8 tháng 4 2017

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\)

\(=\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{10}\right)+\left(\dfrac{1}{11}+...+\dfrac{1}{17}\right)< \dfrac{1}{5}.6+\dfrac{1}{11}.7=\dfrac{6}{5}+\dfrac{7}{11}\)

\(=1\dfrac{46}{55}< 2\)

\(\Rightarrow dpcm\)

23 tháng 7 2017

Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OKleuleu

23 tháng 7 2017

Mình khuyen bạn phải suy nghĩ kĩ bài trước khi đăng lên nhé!!hum

25 tháng 3 2017

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)

\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)

\(\Rightarrow-2\le x\le2\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)

\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)

\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)

\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)

\(\Rightarrow x\in\left\{11;12;13;14\right\}\)

25 tháng 3 2017

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)