Bài 1
a rút gọn B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
b Chứng minh A=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{5}{8}\)
K
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19 tháng 4 2016
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)..................\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.............\frac{19}{20}\)
=\(\frac{1.2.3..............19}{2.3.4..............20}\)
=\(\frac{1}{20}\)
1 tháng 5 2019
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{21}{20}\)
\(B=\frac{21}{2}\)
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KN
1 tháng 5 2019
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{20}\right)\)
\(\Rightarrow B=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{20}{20}+\frac{1}{20}\right)\)
\(\Rightarrow B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{21}{20}\)
\(\Rightarrow B=\frac{21}{2}\)
S
27 tháng 5 2017
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
HN
27 tháng 5 2017
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
21 tháng 7 2015
Ta có:\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}=\frac{1}{20}\)
TN
4 tháng 7 2016
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{19}{20}\)
\(=\frac{1\cdot2\cdot...\cdot19}{2\cdot3\cdot...\cdot20}\)
\(=\frac{1}{20}\)
B= \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)= \(\frac{1}{20}\)
vậy B= \(\frac{1}{20}\)
b,A=(1/101+1/102+...+1/150)+(1/151+1/152+...1/200)>25/125+25/150+25/175+25/200=(1/5+1/6+1/7)+1/8=107/201+1/8>1/2+2/8=5/8
Vậy A>5/8
Nhớ k mik nha!!!!!!!!!!!!!