\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)
\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)
\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)
\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)
b)
\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)
\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)
=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)
\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)
\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)
=> \(B=\frac{140}{69}:2=\frac{70}{69}\)
a) SỬA LẠI ĐỀ : A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{6480}\)
= \(5.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)
=\(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)
= \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)
= \(5.\left(1-\frac{1}{81}\right)\)
= \(5.\frac{80}{81}\)
= \(\frac{400}{81}\)
b) B = \(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+...+\frac{7}{483}\)
= \(\frac{7}{3.5}+\frac{7}{5.7}+\frac{7}{7.9}+...+\frac{7}{21.23}\)
= \(\frac{7}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)
= \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{23}\right)\)
= \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{23}\right)\)
= \(\frac{7}{2}.\frac{20}{69}\)
= \(\frac{70}{69}\)
b> =>B=7*((1/(3*5)+1/(5*7)+....+1/(21*23))
=> B=7*(1/3-1/23)
=>b= 7*20/69
=>B=140/69
VẬY B=140/69
cÂU A SAI ĐỀ BẠN ƠI 6480 CHỨ KO PHẢI 6408
