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\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+...+\dfrac{1}{192}\)
\(2A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{96}\)
\(A=\dfrac{1}{3}-\dfrac{1}{192}=\dfrac{64}{192}-\dfrac{1}{192}=\dfrac{63}{192}=\dfrac{21}{64}\)
-Ta có: \(\dfrac{1}{2}=\dfrac{10}{20}=\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có 10 số \(\dfrac{1}{20}\)).
Mà \(\dfrac{1}{20}< \dfrac{1}{19};\dfrac{1}{20}< \dfrac{1}{18};...;\dfrac{1}{20}< \dfrac{1}{11}\)
\(\Rightarrow\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}< \dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{11}\)
\(\Rightarrow\dfrac{1}{2}< S\)
Để olm.vn giúp em nhá
C = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.11}\)+...+ \(\dfrac{1}{994.997}\) + \(\dfrac{1}{997.1000}\)
C = \(\dfrac{1}{3}\).( \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\)+...+ \(\dfrac{3}{994.997}\)+ \(\dfrac{3}{997.1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+...+ \(\dfrac{1}{994}\)- \(\dfrac{1}{997}\)+ \(\dfrac{1}{997}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\). \(\dfrac{999}{1000}\)
C = \(\dfrac{333}{1000}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(A=1-\dfrac{1}{128}=\dfrac{127}{128}\)
b) ( 1- \(\dfrac{1}{2}\) ) . ( 1 - \(\dfrac{1}{3}\) ) . ( 1 - \(\dfrac{1}{4}\) ) ... ( 1 - \(\dfrac{1}{x}\) ) = 0,01
\(\Rightarrow\) \(\dfrac{1}{2}\) . \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ... \(\dfrac{x-1}{x}\) = 0,01
\(\Rightarrow\) \(\dfrac{1.2.3...\left(x-1\right)}{2.3.4...x}\) = 0,01
\(\Rightarrow\) \(\dfrac{1}{x}\) = 0,01
\(\Rightarrow\) \(\dfrac{1}{x}\) = \(\dfrac{1}{100}\)
\(\Rightarrow\) x = 100
Vậy x =100
lớp 5 đi hỏi 1+1 nhưng hoi bằng 2 nhé