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Ta có :
\(S=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)
\(\Leftrightarrow\)\(3S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Leftrightarrow\)\(3S-S=\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\right)\)
\(\Leftrightarrow\)\(2S=\frac{1}{3}-\frac{1}{3^9}\)
\(\Leftrightarrow\)\(2S=\frac{3^8-1}{3^9}\)
\(\Leftrightarrow\)\(S=\frac{3^8-1}{2.3^9}\)
Ở đây mk chỉ ghi \(...\) cho nhanh nếu bạn làm vào vở thì ghi đầy đủ ra nhé
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}=\frac{31}{32}\)
b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
X : \(3\frac{1}{15}\)=\(1\frac{1}{2}\)
X : \(\frac{46}{15}\)=\(\frac{3}{2}\)
X = \(\frac{3}{2}\)X \(\frac{46}{15}\)
X = \(\frac{23}{5}\)
#)Giải :
\(x\div3\frac{1}{5}=1\frac{1}{2}\)
\(\Leftrightarrow x\div\frac{16}{5}=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{24}{5}\)hay \(4\frac{4}{5}\)
\(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Leftrightarrow\left(\frac{2}{3}-\frac{1}{2}\right)x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{2}\)
a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
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 = \(\frac{\frac{1}{3}^9-\frac{1}{3}}{-\frac{2}{3}}\)