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a) \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> \(D=1-\frac{1}{1024}\)
b) \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)
\(D=1-\frac{1}{1024}\)
\(D=\frac{1023}{1024}\)
\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(Đ=1-\frac{1}{20}\)
\(Đ=\frac{19}{20}\)
Phần c như kiểu sai đề chỗ cuối hay sao ấy.
(1-1/12) x (1-1/11) x (1-1/10) x (1-1/9) x (1-1/8)
=(1-1) x (1/12+1/11+1/10+1/9+1/8)
=0 x (1/12+1/11+1/10+1/9+1/8)
=0
ko bt có đúng ko nữa
\(\left(1-\frac{1}{12}\right).\left(1-\frac{1}{11}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{8}\right)\)
\(=\frac{11}{12}.\frac{10}{11}.\frac{9}{10}.\frac{8}{9}.\frac{7}{8}\)
\(=\frac{11.10.9.8.7}{12.11.10.9.8}\)
\(=\frac{7}{12}\)
\(\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>\dfrac{1}{6}\)
\(\dfrac{1}{5}=\dfrac{11}{55}\)> \(\dfrac{11}{63}\) > \(\dfrac{11}{66}\) = \(\dfrac{1}{6}\)
⇒ \(\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>\dfrac{11}{63}>\dfrac{1}{6}\)
Vậy phân số lớn nhất trong các phân số đã cho là \(\dfrac{1}{3}\)
