a) \(\frac{1}{9}\)

b) -1100

đây là toán lớp 6 sao trông khó khó

    \(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

=>\(B=\frac{1.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{3.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{14}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{\frac{4}{4}-\frac{4}{16}-\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)

=>\(B=\frac{1}{3}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

=>\(B=\frac{1}{3}.\frac{3}{4}+\frac{5}{8}\)

=>\(B=\frac{1}{4}+\frac{5}{8}\)

=>\(B=\frac{2}{8}+\frac{5}{8}\)

=>\(B=\frac{7}{8}\)

l-i-k-e cho mình nhé bạn.

a) Ta có\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\)

b) Ta có \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2048}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)(1)

Đặt S = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)

=> \(2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)

Lấy 2S trừ S ta có :

2S - S \(=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\right)\)

\(S=1-\frac{1}{2048}\)

Khi đó (1) <=> \(1-\left(1-\frac{1}{2048}\right)=1-1+\frac{1}{2048}=\frac{1}{2048}\)

\(a,\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{90}+\frac{2}{110}\)

\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{90}+\frac{1}{110}\right)\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=1-\frac{2}{11}\)

\(=\frac{9}{11}\)

các bn ơi giải giúp mình đi mà

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

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

\(=1-\frac{1}{128}\)

\(\frac{127}{128}\)

127/128