\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

Đặt A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

1093/729

Úi

2A = 3 - \(\dfrac{1}{729}=\dfrac{2186}{729}\) 

A = \(\dfrac{2186}{729}:2=\dfrac{1093}{729}\)

2 điểm!?

thi hay sao?

S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )

S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)

\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)

\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)

b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)

=1/3-1/3

=0

c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

=2016/2017

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

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

\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)

\(\dfrac{1}{3}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)+\(\dfrac{1}{729}\)

=\(\dfrac{243}{729}\)+\(\dfrac{81}{729}\)+\(\dfrac{27}{729}\)+\(\dfrac{3}{729}\)+\(\dfrac{1}{729}\)

=\(\dfrac{355}{729}\)

chúc bạn học tốt ạ

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 ^^

Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)

\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)

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

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

=>A là số hữu tỉ (ĐPCM)