\(C=\left(\frac{1}{2}-1\right)+\left(1-\frac{3}{4}\right)+\left(\frac{7}{8}-1\right)+...+\left(1-\frac{1023}{1024}\right)\)

\(C=\left(\frac{1}{2^1}-\frac{2}{2}\right)+\left(\frac{2^2}{2^2}-\frac{3}{2^2}\right)+...+\left(\frac{1024}{1024}-\frac{1023}{2^{10}}\right)\)

\(C=\frac{-1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(2C=-1+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^9}\)

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

\(3C=\frac{1}{2^{10}}-1\)

\(C=\frac{\frac{1}{2^{10}}-1}{3}\)

hok tốt!!

\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)

Hok tốt

Yume Nguyễn bạn giải giúp mk phần b đc k

ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

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

tách

\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2B-B=\frac{1}{2}-\frac{1}{1024}\)

thay vào B ta có 

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

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)

\(\Rightarrow A=\frac{513}{1024}\)

1) tự làm (thực hiện từ dưới lên)

2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)

      = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)

     = \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)

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

1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)

\(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=\frac{50}{7}\).

B = \(\frac{\frac{1}{2^{10}}.5-\frac{1}{\left(2^2\right)^5}.3}{\frac{1}{2^{10}}.\frac{1}{3}-\frac{1}{2^{11}}}=\frac{\frac{1}{2^{10}}.\left(5-3\right)}{\frac{1}{2^{10}}.\left(\frac{1}{3}-\frac{1}{2}\right)}=\frac{2}{\left(-\frac{1}{6}\right)}=2:\left(-\frac{1}{6}\right)=-12\)