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\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
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
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
\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7\times\frac{6}{70}\)
\(=\frac{6}{10}=\frac{3}{5}\)
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
\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\\ 7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+....+\frac{1}{69}-\frac{1}{70}\right)\\ 7.\left(\frac{1}{10}-\frac{1}{70}\right)\\ 7.\frac{6}{70}=\frac{3}{5} \)
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=3/4.7+3/7.10+...+3/73.76
A=1/4-1/7+1/7-1/10+1/10-1/13+....+1/73-1/76
A=1/4-1/76
A=9/38
b) B=5/10.11+5/11.12+....+5/99.100
B=5(1/10.11+1/11.12+1/12.13+...+1/99.100)
B=5(1/10-1/11+1/11-1/12+1/12-1/13+...+1/99-1/100)
B=5(1/10-1/100)
B=5.99/100
B=99/20
\(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+.......+\frac{1}{19\cdot20}\)\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{18}-\frac{1}{19}\)\(+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)
K CHO MÌNH NHA CÁC BẠN
\(S=\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}=\frac{9}{100}\)
Vì những phần tử còn lại đã tự khử nhau rồi nhé ^^