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A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
a. Ta có :
B = 308/1 + 307/2 +306/3+....+1/308
B = (1+1+....+1) + 307/2 + ....+ 1/308
B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1
B = 309/2 + 309/3 + ....+ 309/308 + 309/309
B = 309.(1/2 + 1/3 + ....+1/309)
Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308
A/B = 1/2 + 1/3 +... + 1/309 / 309.(1/2 + 1/3 + ....+1/309)
A/B = 1/309
b.7/10.11 + 7/11.12 + .... +7 /69.70
= 7. (1/10.11+1/11.12 + ...+ 1/69.70)
= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)
= 7.(1/10 - 1/70)
= 7. 3/35
= 3/5
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7\times\frac{6}{70}\)
\(=\frac{6}{10}=\frac{3}{5}\)
Gọi tổng trên là A
A = 7/10.11 + 7/11.12 +.....+ 7/69.70
A = 7(1/10.11 + 1/11.12 +.....+ 1/69.70)
A =7( 1/10 - 1/11 + 1/11 - 1/12 +.....+ 1/69 - 1/70)
A = 7( 1/10 - 1/70)
A = 7 . 3/35
A = 21/35
#)Giải :
\(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(2B=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(2B=\frac{1}{25}-\frac{1}{75}\)
\(2B=\frac{2}{75}\)
\(B=\frac{2}{75}\div2\)
\(B=\frac{1}{75}\)
a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)
\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)
\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)
\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)
\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)
A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
a,
suy ra A = 7. (1/10.11+1/11.12+1/12.13+.......+1/69.70)
suy ra A = 7. ( 1/10 - 1/11+ 1/11 - 1/12 + 1/12 - 1/13+ ............. + 1/69 - 1/70)
suy ra A = 7. ( 1/ 10 - 1/70)
suy ra A= 7. 3/35
suy ra A= 3/5
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