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2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
\(=>C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}.....\cdot\dfrac{101}{100}\)
\(C=\dfrac{3\cdot4\cdot5.......\cdot101}{2\cdot3\cdot4.........\cdot100}\)
\(C=\dfrac{101}{2}\)
A=\(x.\dfrac{1}{5}+x.\dfrac{2}{3}-x.\dfrac{1}{4}\)
=\(x.\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{4}\right)\)
=\(x.\dfrac{37}{60}\)
Thay x=\(\dfrac{1}{2}\) vào A ta được
A=\(\dfrac{1}{2}.\dfrac{37}{60}=\dfrac{37}{120}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow\dfrac{1}{2}P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)
\(\Rightarrow\dfrac{1}{2}P-P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{100}}\)
\(\Rightarrow-\dfrac{1}{2}P=\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\)
\(\Rightarrow P=\left(\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\right):\left(-\dfrac{1}{2}\right)\)