\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
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a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)
= \(\frac{15}{8}+\frac{5}{8}\)
= \(\frac{5}{2}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)
\(=\frac{49}{34}\)
Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}+\frac{1}{3^9}\)
=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>3A-A=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\)\(\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}+\frac{1}{3^9}\right)\)
=>2A=\(1-\frac{1}{3^9}\)
=>A=\(\frac{9841}{19683}\)
đặt biểu thức trên là A
ta có
3A=3(1/3+1/3^2+1/3^3+1/3^4+....+1/3^9)
3A=1+1/3+1/3^2+...+1/3^8
3A-A=1+1/3+1/3^2+...+1/3^8-(1/3+1/3^2+1/3^3+..+1/3^9)
2A=1-1/3^9
2A=3^9-1/3^9
A=3^9-1/3^9.2
vậy A=3^9-1/3^9.2
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
\(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
=>\(B=\frac{1.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{3.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{14}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{\frac{4}{4}-\frac{4}{16}-\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3}{4}+\frac{5}{8}\)
=>\(B=\frac{1}{4}+\frac{5}{8}\)
=>\(B=\frac{2}{8}+\frac{5}{8}\)
=>\(B=\frac{7}{8}\)
l-i-k-e cho mình nhé bạn.
Đặt A = 1/3 + 1/32 + 1/33 + ... + 1/38
=> 1/3A = 1/3.(1/3 + 1/32 + 1/33 + ... + 1/38)
=> 1/3A = 1/32 + 1/33 + 1/34 + ... + 1/39
=> 1/3A - A = (1/32 + 1/33 + 1/34 + ... + 1/39) - (1/3 + 1/32 + 1/33 + ... + 1/38)
=> -2/3A = 1/39 - 1/3
=> A = \(\frac{\frac{1}{3}^9-\frac{1}{3}}{-\frac{2}{3}}\)