Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)
\(=\dfrac{11}{20}\)
Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)
\(=\dfrac{11}{20}\)
thực hiện phép tính
\(\dfrac{16}{103}\)+(38+\(\dfrac{-16}{103}\))
\(\dfrac{100}{91}\)+310+\(\dfrac{-9}{91}\)
\(\dfrac{-13}{49}\)+(\(\dfrac{-36}{49}\)+41)
\(\dfrac{-10}{71}\):1\(\dfrac{4}{71}\)
\(\dfrac{4}{15}\)+\(\dfrac{8}{15}\):2 -\(\dfrac{1}{18}\).\(\left(-3\right)^2\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
bài 1 tính :
\(\dfrac{-8}{9}\) . \(\dfrac{12}{19}\) . \(\dfrac{9}{-4}\) . \(\dfrac{19}{24}\) \(\dfrac{-5}{16}\) . \(\dfrac{17}{15}\) : \(\dfrac{-17}{8}\)
\(\dfrac{4}{13}\) . \(\dfrac{2}{7}\) + \(\dfrac{-3}{26}\) + \(\dfrac{4}{13}\) . \(\dfrac{5}{7}\) \(\dfrac{6}{11}\) . \(\dfrac{3}{4}\) + \(\dfrac{-12}{60}\) +\(\dfrac{-3}{4}\) .\(\dfrac{-5}{11}\)
Cho \(S=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\). Chứng tỏ rằng S<\(\dfrac{1}{16}\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\) . Chứng tỏ rằng \(S< \dfrac{1}{16}\)
Tính: \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.\dfrac{35}{36}.\dfrac{48}{49}.\dfrac{63}{64}\)
\(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+...+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)<\(\dfrac{3}{16}\)CMR
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
Chứng minh: \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)<\(\dfrac{3}{16}\)