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\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{101}\)
\(\Rightarrow2B-B=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{101}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow B=1-\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
\(\Rightarrow B< 1\)(đpcm)
_Chúc bạn học tốt_
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
1.\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{49}{50}=\frac{1}{50}\)
Ta có: \(D=2016\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2017}\right)\)
\(=2016.\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\)\(=2016.\left(\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\right)\)
\(=2016\left(\frac{1.3.5.7...2015}{3.5.7....2015.2017}\right)\)\(=2016.\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(D=\frac{2016}{2017}\)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+...+100}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{5050}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}....\frac{5049}{5050}=\frac{4}{6}.\frac{10}{12}...\frac{10098}{10100}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{99.102}{100.101}\)
\(A=\frac{1.2...98.99}{2.3...99.100}.\frac{4.5...102}{3.4...101}=\frac{1}{100}.\frac{102}{3}\)
Vậy \(A=\frac{102}{300}=\frac{17}{50}\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+...+100}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{5050}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.....\frac{5049}{5050}\)
\(=\frac{4}{6}.\frac{10}{12}.....\frac{10098}{10100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.....\frac{99.102}{100.101}\)
\(=\frac{1.2.3..98.99}{2.3.4..99.100}.\frac{4.5.6...102}{3.4.5...101}\)
\(=\frac{1}{100}.\frac{102}{3}\)
\(=\frac{17}{50}\)