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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) 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
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
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 \(D=\frac{2016}{2017}\)
1.\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{49}{50}=\frac{1}{50}\)
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
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Câu 1:
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...\left(1+\frac{1}{2014.2016}\right)\)
\(\Rightarrow C=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{2015.2015}{2014.2016}\)
\(\Rightarrow C=\frac{4.9.16...2015.2015}{3.8.15...2014.2016}\)
\(\Rightarrow C=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4...2014.2016}\)
\(\Rightarrow C=\frac{2.3.4...2015.2.3.4...2015}{1.2.3...2014.3.4.5...2016}\)
\(\Rightarrow C=\frac{2015}{1008}.\)
Vậy \(C=\frac{2015}{1008}.\)
Câu 2:
Do p là số nguyên tố lớn hơn 3 nên p có dạng \(3k+1\)hoặc\(3k+2\)
+ Nếu \(p=3k+1\Rightarrow p^2-1=\left(3k+1\right)^2-1\)
\(=9k^2+3k+3k+1-1\)
\(=9k^2+6k⋮3.\)( 1 )
+ Nếu \(p=3k+2\Rightarrow p^2-1=\left(3k+2\right)^2-1\)
\(=9k^2+6k+6k+4-1\)
\(=9k^2+12k+3⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow p^2-1⋮3\left(đpcm\right).\)
Câu 3:
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1000^{10}=10^{30}.\)( 1 )
\(2^{100}=2^{31}.2^6.2^{63}=2^{31}.64.512^7< 2^{31}.125.625^7=2^{31}.5^{31}=\)\(10^{31}.\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow10^{30}< 2^{100}< 10^{31}.\)
\(\Rightarrow\)2100 khi viết trong hệ thập phân có 31 chữ số.
Đáp số: 31 chữ số.
Câu 1 :
C = (1 + 1/1.3)(1 + 1/2.4)(1 + 1/3.5) .... (1 + 1/2014.2016)
C = (1.3/1.3 + 1/1.3) (2.4/2.4 + 1/2.4) ... (2014.2016/2014.2016 + 1/2014.2016)
C = 2.2/1.3 * 3.3/2.4 * ... * 2015.2015/2014.2016
C = 2.3....2015/1.2....2014 * 2.3....2015/3.4....2016
C = 2015 * 1/1008
C = 2015/1008
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+...+100}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{5050}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}....\frac{5049}{5050}=\frac{4}{6}.\frac{10}{12}...\frac{10098}{10100}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{99.102}{100.101}\)
\(A=\frac{1.2...98.99}{2.3...99.100}.\frac{4.5...102}{3.4...101}=\frac{1}{100}.\frac{102}{3}\)
Vậy \(A=\frac{102}{300}=\frac{17}{50}\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+...+100}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{5050}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.....\frac{5049}{5050}\)
\(=\frac{4}{6}.\frac{10}{12}.....\frac{10098}{10100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.....\frac{99.102}{100.101}\)
\(=\frac{1.2.3..98.99}{2.3.4..99.100}.\frac{4.5.6...102}{3.4.5...101}\)
\(=\frac{1}{100}.\frac{102}{3}\)
\(=\frac{17}{50}\)