A=1/3*7+1/7*11+..+1/95*99
=> 4A=4/3*7+4/7*11+..+4/95*99
=>4A=1/3-1/7+1/7-1/11+...+1/95-1/99=1/3-1/99=32/99
=>A=8/99
\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)
\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)
\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}\cdot\frac{32}{99}=\frac{8}{99}\)
Đặt A=\(\frac{1}{3.7}\)+\(\frac{1}{7.11}\)+\(\frac{1}{11.15}\)+.....+\(\frac{1}{95.99}\)
4A=\(\frac{7-3}{3.7}\)+\(\frac{11-7}{7.11}\)+\(\frac{15-11}{15.11}\)+...+\(\frac{99-95}{95\cdot99}\)
=\(\frac{7}{3.7}\)-\(\frac{3}{3.7}\)+\(\frac{11}{7.11}\)-\(\frac{7}{7.11}\)+\(\frac{15}{15.11}\)-\(\frac{11}{15.11}\)+...+\(\frac{99}{95.99}\)-\(\frac{95}{95.99}\)
=\(\frac{1}{3}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{15}\)+...+\(\frac{1}{95}\)-\(\frac{1}{99}\)
=\(\frac{1}{3}\)-\(\frac{1}{99}\)
=\(\frac{32}{99}\)
A=\(\frac{8}{99}\)
=(4/3.7+4/7.11+4/11.15+.....+4/95.99).1/4
=(1/4-1/7+1/7-1/11+1/11-1/15+.....+1/95-1/99).1/4
=(1/4-1/99).1/4
=95/396 .1/4=95/1584
Đáp án : 8 / 99
Mk xl vì sự bất tiện này nhé , nếu bạn cần bài giải đầy đủ í , có thể ib mk sẽ giử ảnh nhé
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