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25 tháng 4 2018
Ta có :
\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)
\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)
\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)
\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)
\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)
Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra :
\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{A}{100}=98-\frac{49}{100}\)
\(\frac{A}{100}=\frac{9751}{100}\)
\(A=\frac{9751}{100}.100\)
\(A=9751\)
Vậy \(A=9751\)
Chúc bạn học tốt ~
MT
0
8 tháng 2 2016
A= 5.(1/2 + 1/6+1/12+1/20+...+1/9506+1/9702+1/9900)
= 5. (1/1.2 + 1/2.3+1/3.4+1/4.5+...1/97.98+1/98.99+1/99.100)
= 5 .(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/97-1/98+1/98-1/99+1/99-1/100)
= 5.(1-1/100)=5. 99/100=99/20
6 tháng 3 2016
A=\(\frac{1}{2}\)+\(\frac{5}{6}\)+\(\frac{11}{12}\)+\(\frac{19}{20}\)+\(\frac{29}{30}\)+\(\frac{41}{42}\)=(1-\(\frac{1}{2}\))+(1-\(\frac{1}{6}\))+(1-\(\frac{1}{12}\))+(1-\(\frac{1}{20}\))+(1-\(\frac{1}{30}\))+(1-\(\frac{1}{42}\))
=1-1+\(\frac{1}{2}\)+1-\(\frac{1}{2}\)+\(\frac{1}{3}\)+1-\(\frac{1}{3}\)+\(\frac{1}{4}\)+1-\(\frac{1}{4}\)+\(\frac{1}{5}\)+1-\(\frac{1}{5}\)+\(\frac{1}{6}\)+1-\(\frac{1}{6}\)+\(\frac{1}{7}\)
=(1-1+1+1+1+1+1)+(\(\frac{1}{2}\)-\(\frac{1}{2}\))+(\(\frac{1}{3}\)-\(\frac{1}{3}\))+(\(\frac{1}{4}\)-\(\frac{1}{4}\))+(\(\frac{1}{5}\)-\(\frac{1}{5}\))+(\(\frac{1}{6}\)-\(\frac{1}{6}\))+\(\frac{1}{7}\)
=5+\(\frac{1}{7}\)=\(\frac{36}{7}\)
Hơi nhầm nè , để tôi sửa lại đề \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\)
\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(A=1+1+1+...+1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-....-\frac{1}{9900}\)
\(A=98-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\right)\)
\(A=98-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(A=98-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=98-\left(\frac{1}{2}-\frac{1}{100}\right)=98-\frac{49}{100}=\frac{9751}{100}\)
Vậy.............
\(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9989}{9900}\)
\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
có 50 số 1
\(A=50-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
Đặt B = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
Thay B vào A ta được:
\(A=50-\frac{49}{100}=\frac{5000}{100}-\frac{49}{100}=\frac{4951}{100}\)