Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=2-\left(\frac{2^3}{25}+\frac{2^3}{63}+...+\frac{2^3}{255}+\frac{2^3}{323}\right)\)
\(=2-4.\left(\frac{2}{35}+\frac{2}{63}+...+\frac{2}{255}+\frac{2}{323}\right)\)
\(=2-4.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{15.17}+\frac{2}{17.19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\right)\)
\(=2-4.\left(\frac{1}{5}-\frac{1}{19}\right)\)
\(=2-4.\frac{14}{95}=2-\frac{56}{95}=\frac{134}{95}\)
Đặt A=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) +\(\frac{3}{3^3}\) - \(\frac{4}{3^4}\)+...+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3A=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\)+...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4A=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)- \(\frac{100}{3^{100}}\)
=> 4A<1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\) (1)
Đặt B=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)
=> B=2+ \(\frac{1}{3}\) - \(\frac{1}{3^2}\) +...+\(\frac{1}{3^{97}}\) - \(\frac{1}{3^{98}}\)
=> 4B=B+3B=3-\(\frac{1}{3^{99}}\)<3 => A<\(\frac{3}{4}\) (2)
Từ (1) và (2) ta có: 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
S = \(3+3.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2A = \(1+\frac{1}{2}+...+\frac{1}{2^8}\)
=> 2A - A = A = \(\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=1-\frac{1}{2^9}\)
=> S = 3 + 3 . A = \(3+3.\left(1-\frac{1}{2^9}\right)=3+3-\frac{3}{2^9}=6-\frac{3}{2^9}\)
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
