Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow2A-A=1-\frac{1}{32}\)
\(\Rightarrow A=\frac{31}{32}\)
1/2+1/4+1/8+1/16+1/32=1/2+(1/2-1/4)+(1/4-1/8)+(1/16-1/32)
=1/2-1/32
=15/32
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
4-\(\frac{4}{9}\)- [\(2\frac{1}{4}\)+1\(\frac{4}{9}\)]
=\(4-\frac{4}{9}-2\frac{1}{4}-1\frac{4}{9}\)
=\(\left(4-2\frac{1}{4}\right)-\left(\frac{4}{9}-1\frac{4}{9}\right)\)
=\(1\frac{3}{4}-\left(-1\right)\)
=\(2\frac{3}{4}\)
= 1-1,25+1+0,75
=( 1+ 1 ) - 1,25 +0,75
=2 - 1,25 + 0,75
= 0,75+0,75
=1,5
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot....\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot....\cdot2003}{2\cdot3\cdot....\cdot2004}\)
\(=\frac{1}{2004}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{1258}+\frac{1}{1480}\)
\(=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{34\times37}+\frac{1}{37\times40}\)
\(=\frac{1}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{34\times37}+\frac{3}{37\times40}\right)\)
\(=\frac{1}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{34}-\frac{1}{37}+\frac{1}{37}-\frac{1}{40}\right)\)
\(=\frac{1}{3}\times\left(1-\frac{1}{40}\right)=\frac{1}{3}\times\frac{39}{40}=\frac{13}{40}\)
1/4+1/28+1/70+...+1/1258+1/1480
=1/1.4+1/4.7+1/7.10+...+1/3437+1/37.40
=1/3(1-1/4+1/4-1/7+1/7-1/10+...+1/34-1/37+1/37-1/40)
=1/3(1-1/40)
=1/3.39/40
=13/40