\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(< =>\frac{1}{1}-\frac{1}{100}\)

\(< =>\frac{99}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))

A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128

A=1-1/128

A=127/128

Vậy A=\(\frac{127}{128}\)

B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

B=1-1/100

B=99/100

Vậy B=\(\frac{99}{100}\)

a.1025

mk chịu câu b

*** mk nha

a) = \(\frac{127}{96}\)

b) = \(\frac{255}{256}\)

c) Mik bỏ nha

d) = \(\frac{1023}{512}\)

e) = \(\frac{2343}{625}\)

bạn có thể trả lời rõ ra được ko

a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)

\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)

\(\frac{1}{4}xA=\frac{127}{384}\)

\(A=\frac{127}{384}:\frac{1}{4}\)

\(A=\frac{127}{96}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

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