Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^9}\)
2A = \(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^8}\)
2A - A = \(1-\frac{1}{2^9}\)
=> A = \(1-\frac{1}{2^9}\)
2A = 1 + 1/2 + 1/2^2 + ... + 1/2^8
2A - A = 1 + 1/2 + .. + 1/2^8 - 1/2 - 1/2^2 - .. - 1/2^9
A = 1 - 1/2^9
A = 1 - 1/512
A = 511/512
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Ta có :
\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)
\(\Rightarrow A-\frac{1}{2}A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2^{10}}\)
\(\Rightarrow A=1-\frac{1}{2^9}=\frac{511}{512}\)