mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

a)hình như đề sai thì phải

sửa lại

\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)

=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)

=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)

a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)

\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)

c,

\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)

\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)

d,

\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)

Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)

...

\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)

\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A< 1-\dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{9}{10}\)

\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))

$\genfrac{}{}{0ex}{}{1}{2^2}=\genfrac{}{}{0ex}{}{1}{2\cdot 2}<\genfrac{}{}{0ex}{}{1}{1\cdot 2}$

$\genfrac{}{}{0ex}{}{1}{3^2}=\genfrac{}{}{0ex}{}{1}{3\cdot 3}<\genfrac{}{}{0ex}{}{1}{2\cdot 3}$

$\genfrac{}{}{0ex}{}{1}{4^2}=\genfrac{}{}{0ex}{}{1}{4\cdot 4}<\genfrac{}{}{0ex}{}{1}{3\cdot 4}$

...

$\genfrac{}{}{0ex}{}{1}{9^2}=\genfrac{}{}{0ex}{}{1}{9\cdot 9}<\genfrac{}{}{0ex}{}{1}{8\cdot 9}$

$\genfrac{}{}{0ex}{}{1}{10^2}=\genfrac{}{}{0ex}{}{1}{10\cdot 10}<\genfrac{}{}{0ex}{}{1}{9\cdot 10}$

$\Rightarrow A=\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+\genfrac{}{}{0ex}{}{1}{4^2}+...+\genfrac{}{}{0ex}{}{1}{10^2}<\genfrac{}{}{0ex}{}{1}{1\cdot 2}+\genfrac{}{}{0ex}{}{1}{2\cdot 3}+\genfrac{}{}{0ex}{}{1}{3\cdot 4}+...+\genfrac{}{}{0ex}{}{1}{9\cdot 10}$

$\Rightarrow A<1-\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{3}+...+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{10}$

$\Rightarrow A<1-\genfrac{}{}{0ex}{}{1}{10}$

$\Rightarrow A<\genfrac{}{}{0ex}{}{9}{10}$

$\Rightarrow A<1$ (vì: $\genfrac{}{}{0ex}{}{9}{10}<1$)

Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)

\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)

\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)

\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)

Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)

\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.

Vậy \(\dfrac{B}{A}\) là số nguyên.

A=1/5^3+1/6^3+...+1/2023^3

1/5^3<1/4*5*6

Xét tương tự, ta đều sẽ được:

\(\dfrac{1}{n^3}< \dfrac{1}{n\left(n-1\right)\left(n+1\right)}\)

=>\(A< \dfrac{1}{4\cdot5\cdot6}+\dfrac{1}{5\cdot6\cdot7}+...+\dfrac{1}{2022\cdot2023\cdot2024}\)

=>\(A< \dfrac{1}{2}\left(\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+...+\dfrac{2}{2022\cdot2023\cdot2024}\right)\)

=>\(A< \dfrac{1}{2}\left(\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}+...+\dfrac{1}{2022\cdot2023}-\dfrac{1}{2023\cdot2024}\right)\)

=>A<1/40

Ta có BĐT: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}< \dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)\cdot k\cdot\left(k+1\right)}\)

Do đó, ta được:

\(\dfrac{1}{5\cdot6\cdot7}+\dfrac{1}{6\cdot7\cdot8}+...+\dfrac{1}{2023\cdot2024\cdot2025}< A\)

\(\Leftrightarrow A>\dfrac{1}{2}\left(\dfrac{1}{5\cdot6}-\dfrac{1}{2024\cdot2025}\right)>\dfrac{1}{2}\left(\dfrac{1}{30}-\dfrac{1}{390}\right)=\dfrac{1}{65}\)

=>1/65<A<1/40

Ta có:

\(\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)

\(=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)

\(=1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{1}{2016}\right)\)

\(=\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2016}+\dfrac{2017}{2017}\)

\(=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)

Do đó: \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\right)}=\dfrac{1}{2017}\)

Vậy...