Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sửa đề: A=1/3+1/9+1/27+...+1/6561
=1/3+1/3^2+1/3^3+...+1/3^8
=>3A=1+1/3+...+1/3^7
=>3A-A=1-1/3^8
=>\(2A=\dfrac{3^8-1}{3^8}\)
=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)
Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)
\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)
\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)
\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)
\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)
\(2S=\dfrac{4132}{6561}\)
\(S=\dfrac{2066}{6561}\)
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
\(S=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{2019}{3^{2019}}\)
\(3S=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{2019}{3^{2018}}\)
\(\Rightarrow3S-S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2018}}-\dfrac{2019}{3^{2019}}\)
\(\Rightarrow2S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2018}}-\dfrac{2019}{3^{2019}}\)
\(\Rightarrow6S=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2017}}-\dfrac{2019}{3^{2018}}\)
\(\Rightarrow4S=3-\dfrac{2020}{3^{2018}}+\dfrac{2019}{3^{2019}}=3-\dfrac{1347}{3^{2018}}< 3\)
\(\Rightarrow S< \dfrac{3}{4}\)
Mình không nghĩ TH làm được bài này đâu nên mình làm cách THCS nha, dù sao đây cũng là toán lớp 6 mà!!!
Gọi tổng là A ta có :
\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{6561}\)
\(A=\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\)
\(3A=3.\left(\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\right)\)
\(3A=\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+...+\frac{2}{3^9}\)
\(3A-A=\left(\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+...+\frac{2}{3^9}\right)-\left(\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\right)\)
\(2A=\frac{2}{3^9}-\frac{2}{3^1}\)
\(A=\frac{\frac{2}{3^9}-\frac{2}{3^1}}{2}\)
Vậy,...
Nếu sai mong bạn thông cảm nha!!!