Tính Nhanh:
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)\
Giúp mk nha, cần gấp. Ai nhanh đúng tik cho
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17 tháng 9 2017
\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=1\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=1-\frac{1}{6}\)
\(C=\frac{5}{6}\)
11 tháng 9 2018
từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
26 tháng 7 2017
đơn giản :
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+........+\(\frac{1}{99.100}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
A=1 - \(\frac{1}{100}\)
A= \(\frac{99}{100}\)
CÓ AI DÙNG HỌC 24 GIỜ KO
26 tháng 7 2017
A = 1/2 + 1/6 / + 1/ 12 + 1/20 + ......+ 1/(99.100)
A= 1/ ( 1 x 2 ) + 1/ ( 2 x 3 ) + 1 / ( 3 x 4 ) + .....+ 1/ ( 99 x 100 )
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .................+ 1/99 - 1/100
A= 1 - 1/100
A= 99/100
CHÚC BẠN HỌC TỐT
NK
9 tháng 5 2019
\(D=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)\)
\(=\left(1+1+1+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
\(=4-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}\right)\)
\(=4-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=4-\left(1-\frac{1}{5}\right)=4-\frac{4}{5}=\frac{16}{5}\)
9 tháng 5 2019
\(D=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}\)
\(D=\left(1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\right)\)
\(D=4-\frac{4}{5}\)
\(D=\frac{16}{5}\)
NX
31 tháng 7 2018
\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
16 tháng 4 2018
Bài 1:
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) ta có: \(A=1+2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2019}\)
\(\Rightarrow2A-A=2^{2019}-2\)
\(\Rightarrow A=2^{2019}-2\)
Chúc bn học tốt !!!!!
16 tháng 4 2018
a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)