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Ta có: \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^8}\)
\(\Rightarrow A=\left(1-\frac{1}{3^8}\right)\div2\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^8\times2}\)
A=1/3+1/3^2+1/3^3+...+1/3^8
3A=1+1/3+1/3^2+...+1/3^7
3A-A=1-1/3+1/3-1/3^2+1/3^2-1/3^3+...+1/3^7-1/3^8
2A=1-1/3^8
2A=6560/6561
Suy ra A=3280/6561
nho k cho minh voi nhe
A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
3A = \(1+3+\frac{1}{3^2}+...+\frac{1}{3^7}\)
2A = 3A - A = \(1-\frac{1}{3^8}\)
=> A = \(\frac{1-\frac{1}{3^8}}{2}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>3A=\(1+\frac{1}{3}+...+\frac{1}{3^7}\)
=>3A-A=\(\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
=>2A=\(1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=>A=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)
A=1/3+1/32+1/33+...+1/38
=>3A=1+1/3+1/32+...+1/37
=>3A-A=(1+1/3+1/32+...+1/37)-(1/3+1/32+1/33+...+1/38)
=>2A=1+1/3+1/32+...+1/37-1/3-1/32-1/33-...-1/38
=1-1/38
=\(\frac{6550}{6561}\)
=>A=\(\frac{6560}{6561}:2=\frac{6560}{6561}.\frac{1}{2}=\frac{3280}{6561}\)
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+........+\frac{1}{3^8}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}\)
\(3A-A=1-\frac{1}{3^8}\)
\(A=\left(1-\frac{1}{3^8}\right):2\)
2.9 . 3/-11 + 2/9 . -8/11 + 1/3
= 2/9 . (3/-11 + -8/11) + 1/3
= 2/9 . (-1) + 1/3
= -2/9 + 1/3
= 1/9
3/5 . 7/8 - 3/5 . -1/5
= 3/8 . (7/8 - 1/5)
= 3/8 . 43/40
= 129/320
\(\frac{2}{9}\) x \(\frac{3}{-11}\) +\(\frac{2}{9}\) x\(\frac{-8}{11}\) +\(\frac{1}{3}\) =\(\frac{1}{9}\) \(\frac{3}{5}\) *\(\frac{7}{8}\) -\(\frac{3}{5}\)*\(\frac{-1}{5}\) =\(\frac{129}{200}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\)
\(=>3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)
\(=>3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(=>2A=1-\frac{1}{3^8}=>A=\left(1-\frac{1}{3^8}\right):2\)