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ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
**** cho mh nha bn . thanks
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}...-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)
\(A=1-0+0+0+...+0+0-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
Tính nhanh:
\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)
\(=\left(\frac{1}{1}+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{8}\right)\)\(+\left(\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{4}+\frac{1}{6}\right)+\frac{1}{5}\)
\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{5}\)
\(=\frac{4}{10}+\frac{2}{5}=\frac{2}{5}+\frac{1}{5}=\frac{3}{5}\)
tks giúp mk nha! cảm ơn nhiều ạ...
Đặt \(A=2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=2-\frac{1}{9}=\frac{18}{9}-\frac{1}{9}=\frac{17}{9}\)
b) 1/3+1/3^2+1/3^3+1/3^4+1/3^5 (goi tong bang M)
3M=1+1/3+1/3^2+1/3^3+1/3^4
3M-M=1-1/3^5
2M=242/243
M=242/243*1/2=121/243
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)
Sửa lại là 1/256 nha
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{128}-\frac{1}{256}\right)\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)
\(=1-\frac{1}{256}=\frac{255}{256}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{128}-\frac{1}{256}\right)\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)
\(=1-\frac{1}{256}=\frac{255}{256}\)