b:3=1/2+1/4+1/8+1/16+1/32+1/64+1/128

[b:3]x2=1+1/2+1/4+1/8+1/16+1/32+1/64

[b:3]x2-[b:3]=1+1/2+1/4+1/8+1/16+1/32+1/64-1/2+1/4+1/8+1/16+1/32+1/64+1/128

b:3=1-1/128

b:3=127/128

b=127/128x3

b=381/128

B.2=\(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\)\(+\frac{3}{128}\)

B.2-B=\(\left(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\).\(\left(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\)

B=3-\(\frac{3}{128}\)

B=\(\frac{381}{128}\)

b:3=1/2+1/4+1/8+1/16+1/32+1/64+1/128

[b:3]x2=1+1/2+1/4+1/8+1/16+1/32+1/64

[b:3]x2-[b:3]=1+1/2+1/4+1/8+1/16+1/32+1/64-1/2+1/4+1/8+1/16+1/32+1/64+1/128

b:3=1-1/128

b:3=127/128

b=127/128x3

b=381/128

cho mình 

nhầm tớ lộn sang bài khác sorry

trình bày cách giải giùm với nhé

a: \(=3\cdot\left(\dfrac{1}{4}-\dfrac{6}{7}+\dfrac{8}{21}\right)\)

\(=3\cdot\left(\dfrac{21}{84}-\dfrac{72}{84}+\dfrac{32}{84}\right)\)

\(=\dfrac{-19}{28}\)

b: \(=\dfrac{-2}{3}\left(\dfrac{1}{9}-\dfrac{1}{6}-\dfrac{1}{11}\right)\)

\(=\dfrac{-2}{3}\cdot\dfrac{-29}{198}=\dfrac{29}{99\cdot3}=\dfrac{29}{297}\)

c: \(=\dfrac{-3}{7}+\dfrac{4}{25}+\dfrac{5}{16}+\dfrac{3}{16}\)

\(=\dfrac{-75+28}{175}+\dfrac{1}{2}\)

\(=\dfrac{-47}{175}+\dfrac{1}{2}=\dfrac{-94+175}{350}=\dfrac{81}{350}\)

d: \(=\dfrac{-4}{9}\cdot\left(\dfrac{1}{26}-\dfrac{1}{2}-\dfrac{1}{8}\right)\)

\(=\dfrac{-4}{9}\cdot\dfrac{-61}{104}=\dfrac{61}{26\cdot9}=\dfrac{61}{234}\)