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9 tháng 3 2018
a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}\)
\(=\dfrac{3}{20}\)
c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
NL
0
HT
12 tháng 7 2015
a,\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
b,\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008
QN
7 tháng 3 2018
\(B=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{19.20}\)
\(B=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{19.20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{20}\right)\)
\(B=5.\frac{1}{20}=\frac{1}{4}\)
QN
7 tháng 3 2018
\(C=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\)
\(4C=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(4C=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\)
\(4C=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)
\(4C=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
\(C=\frac{4}{25}:4=\frac{1}{25}\)
HA
Hoàng Anh Sơn
VIP
17 tháng 7
12.A= 1.5.12+5.9(13-1)+9.13(17-5)+13.17(21-9)+.....+97.101(105 - 93)
12.A = 1.5.12 + 5.9.13 -1.5.9 + 9.13.17- 5.9.13 +.....+ 97.101.105 -93.97.101
12.A = 1.5.12 -1.5.9 + 97.101.105
A = (1.5.12 -1.5.9 + 97.101.105):12 = 85725
DT
0
DT
0
DT
0
2 tháng 8 2015
a ) tổng 10 số hạng dầu tiên là :
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/10.11
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/10 - 1/11
= 1/1 - 1/11
= 10/11
28 tháng 2 2017
phan a la10/11 dung 100%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ta có
(X-1):4+1=50
=>X=195
Gọi tổng đó là S
S=42 /1.5+42/5.9+....+42/193.195
S=4(1-1/5+1/5-1/9+...+1/193-1/195)
S=4(1-1/195)
=> S=776/195
mik nha