\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)

\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)

\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))

\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)

\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)

\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)

\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)

Lời giải:

Ta có:

\(A=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2017}{4^{2017}}\)

\(\Rightarrow 4A=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\)

Lấy vế sau trừ vế trước:

\(\Rightarrow 3A=1+\frac{2-1}{4}+\frac{3-2}{4^2}+\frac{4-3}{4^3}+...+\frac{2017-2016}{4^{2016}}-\frac{2017}{4^{2017}}\)

\(\Leftrightarrow 3A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\)

\(\Rightarrow 12A=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\)

Lấy vế sau trừ vế trước suy ra:

\(9A=4-\frac{2017}{4^{2016}}-\frac{1}{4^{2016}}+\frac{2017}{4^{2017}}\)

\(9A=4-\frac{2018}{4^{2016}}+\frac{2017}{4^{2017}}<4-\frac{2018}{4^{2016}}+\frac{2017}{4^{2016}}=4-\frac{1}{4^{2016}}<4\)

Do đó: \(A< \frac{4}{9}< \frac{4}{8}=\frac{1}{2}\) (đpcm)

\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)

\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)

\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)

\(C=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)

\(\Rightarrow4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=1+\dfrac{1-\dfrac{1}{4^{2016}}}{3}-\dfrac{2017}{4^{2016}}\)

\(\Rightarrow C=\dfrac{1}{3}+\dfrac{1-\dfrac{1}{4^{2016}}}{9}-\dfrac{2017}{4^{2016}.3}< \dfrac{1}{2}\)

Vậy...

Ta có:

\(C=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)

\(4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\)

\(\Rightarrow3C=4C-C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)Đặt \(D=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}\)

\(\Rightarrow4D=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2015}}\)

\(\Rightarrow3D=4D-D=4-\dfrac{1}{4^{2016}}\)

Ta có : Do \(4-\dfrac{1}{4^{2016}}< 4\Rightarrow3D< 4\)

\(\Rightarrow D< \dfrac{4}{3}\)

Thay giá trị tìm được của D vào 3C được:

\(3C=4-\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Do \(D< \dfrac{4}{3}\Rightarrow D-\dfrac{2017}{4^{2017}}< \dfrac{4}{3}\)

Hay \(3C< \dfrac{4}{3}\Rightarrow C< \dfrac{4}{9}< \dfrac{1}{2}\Rightarrow C< \dfrac{1}{2}\left(đpcm\right)\)

Vậy....

tik mik nha !!!

Mk thử giải lại xem có nhớ thôi chứ ko có ý j:v

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

\(4C=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2017}{4^{2016}}\)

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

\(3C=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt:

\(\Rightarrow D=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2016}}\)

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

\(4D=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4D-D=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3D=4-\dfrac{1}{4^{2016}}\)

\(D=\dfrac{4}{3}-\dfrac{1}{4^{2016}.3}\)

Thay vào ta có:
\(3C=\dfrac{4}{3}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(3C< \dfrac{4}{3}\Rightarrow C< \dfrac{4}{9}\)

\(\Rightarrow C< \dfrac{1}{2}\)

Dạng bài tương tự như bài này, bạn áp dụng cách làm vào làm bài của bạn nhé: Câu hỏi của Dao Dao - Toán lớp 7 | Học trực tuyến

\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2018}}{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}\)

Đặt \(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\) là B

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

\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2018}}{2018\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}{2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017}{2018}\)

Mơn cac ban, minh dang can cau nay ^^