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a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)
A =\(1-\frac{1}{2^7}\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=1-\frac{1}{11}\)
\(A=\frac{10}{11}\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)
\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\left(2\right)\)
Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Rightarrow B=1-\frac{1}{2^7}\)
\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)
HOK TOT
S = \(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{17x19}\)
2S = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\)\(\frac{1}{17}-\frac{1}{19}\)
2S = \(\frac{1}{3}-\frac{1}{19}\)
2S = \(\frac{16}{57}\)
S = \(\frac{16}{57}\times\frac{1}{2}\)
S = \(\frac{8}{57}\)
\(S=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}+\frac{1}{255}+\frac{1}{323}\)
\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}\)
\(2S=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}\)
\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\)
\(2S=\frac{1}{3}-\frac{1}{19}\)
\(2S=\frac{19}{57}-\frac{3}{57}\)
\(2S=\frac{16}{57}\)
\(S=\frac{16}{57}:2\)
\(S=\frac{16}{57}\cdot\frac{1}{2}\)
\(S=\frac{8}{57}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9+\left(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+...+\frac{1}{9\cdot10}\right)\)
\(A=9+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9+\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
Tìm x biết:
\(\frac{x}{3}-\frac{3}{4}=\frac{1}{12}\)
\(\frac{x}{3}=\frac{1}{12}+\frac{3}{4}\)
\(\frac{x}{3}=\frac{5}{6}\)
\(x=\frac{5}{6}.3\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{199}{230}\)
\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
Vậy \(x=\frac{3}{10}\)
Bài 2: tính
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{5}-\frac{1}{11}\)
\(=\frac{6}{55}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Bài 2:
1/30+1/42+1/56+1/72+1/90+1/110
=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11
=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1/5-1/11=6/55
b)1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50
=49/50
\(A=10.\left(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{71}{72}+\frac{89}{90}\right)\)
Đặt \(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{71}{72}+\frac{89}{90}\)
\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(B=1+1+1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8,1\)
Ta có \(A=10.B=10.B=10.8,1=81\)
Vậy \(A=81\)
A=1/2+1/6+1/12+1/20+1/30
A=1/1x2+1/2x3+1/3x4+1/3x5+1/5x6
A=1/1-1/2+1/2-1/3+.......+1/5-1/6
A=1/1-1/6
A=5/6
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
a) = 1/5.6 + 1/6.7 + 1/7.8 + .. + 1/10.11
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11
= 1/5 - 1/11
= 6/55
b) 2/35 + 2/63 + 2/99 + .. + 2/255
= 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15 + 2/15.17
= 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 + 1/15 - 1/17
= 1/5 - 1/17
= 12/85