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đặt S=\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=>3S= \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
=>3S-S=\(\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
=>s=1-1/729 = 728/729
1/3+1/9+1/27+1/81+1/243+1/729=(1/3+1/9+1/81)+(1/27+1/243+1/729)=37/81+37/729=333/729+37/729=370/729
bài 1 tính nhanh
mik xin sửa đề câu a thành thế này ~
\(a,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2-A=\) ( \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) ) - ( \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(B\cdot3-B=\) ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) )
\(B\cdot2=\) \(1-\frac{1}{729}\)
\(B\cdot2=\frac{728}{729}\)
\(B=\frac{728}{729}:2\)
\(B=\frac{364}{729}\)
\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=\frac{1}{1}-\frac{1}{6}\)
\(C=\frac{5}{6}\)
Gọi S là tổng của biểu thức:
\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}.\)
\(3S=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3S-S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^6}\)
\(2S=1-\frac{1}{3^6}\Rightarrow S=\left(1-\frac{1}{3^6}\right):2\)
Tổng = 243/729 + 81/729 + 9/729 + 3/729 + 1/729
= (243+81+9+3+1)/729
= 337/729
1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
=1+ 243/729+ 81/729 + 27/729 + 9/729 + 3/729
=1093/729
e quy đồng ra nháp rồi cộng các số phù hợp là ra kết quả
A=1+1/3+1/3^2+......1/3^6
3A= 3 +1 + 1/3+......=1/3^5
3A-A= 3-1/3^6
A=\(\frac{3-\frac{1}{3^6}}{2}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
