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A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
32A = \(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
9A - A = \(\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
8A = \(9-\frac{1}{3^{100}}=9-\frac{1}{3^n}\)
=> n = 100
\(\frac{x}{5}=\frac23\)
\(x\) = \(\frac23\times5\)
\(x=\frac{10}{3}\)
Vậy \(x=\frac{10}{3}\)
\(\frac{x}{3}-\frac12=\frac15\)
\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)
\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(x=\frac{7}{10}\times3\)
\(x=\frac{21}{10}\)
Vậy \(x=\frac{21}{10}\)
\(\frac{x}{5}+\frac12=\frac{6}{10}\)
\(\frac{x}{5}=\frac{6}{10}-\frac12\)
\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)
\(\frac{x}{5}=\frac{1}{10}\)
\(x=\frac{1}{10}\times5\)
\(x=\frac12\)
Vậy \(x=\frac12\)
\(\frac{x+3}{15}\) = \(\frac13\)
\(x+3=\frac13\times15\)
\(x+3=5\)
\(x=5-3\)
\(x=2\)
Vậy \(x=2\)
de ot
9A-A=(\(9+1+\frac{1}{3^2}+...+\frac{1}{3^{99}}\))-\(\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
8A=\(9-\frac{1}{3^{100}}\)
=>n=100
a) = 3/3 x ( -24/54 +45/54 ) : 7/12
= 1 x 21/54 x 12/7
= 18/27
( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )
S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)
S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)
S = \(\frac12\) - \(\frac19\)
S = \(\frac{9}{18}-\frac{2}{18}\)
S = \(\frac{7}{18}\)
Bài 3:
Gọi tử là x
Theo đề, ta có: \(\dfrac{x+16}{35}=\dfrac{x}{7}\)
=>7x+112=35x
=>x=4
Nhầm
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3^2A=3^2+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
=> n = 100
Ta có: \(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
\(\Rightarrow9-\frac{1}{3^{100}}=9-\frac{1}{3^n}\)
\(\Rightarrow\frac{1}{3^{100}}=\frac{1}{3^n}\)
\(\Rightarrow3^{100}=3^n\)
\(\Rightarrow n=100\)
Vậy n = 100
\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3^2A=3^2+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\)
\(\Rightarrow9A-A=\left(9+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow8A=9-\frac{1}{3^{100}}\)
\(\Rightarrow n=100\)
giúp nha, mình gấp lắm
n=100