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1/1024 câu này trên violimpic vòng 2 và mình làm đúng rồi
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(\Rightarrow A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
\(2A+A=\left(2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)+\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(\Rightarrow3A=2-\frac{1}{1024}\)
\(\Rightarrow3A=\frac{2048}{1024}-\frac{1}{1024}\)
\(\Rightarrow3A=\frac{2047}{1024}\)
\(\Rightarrow A=\frac{2047}{1024}:3\)
\(\Rightarrow A=\frac{2047}{3072}\)
gọi A=\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
2xA=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
2xA‐A=﴾1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)﴿‐﴾\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)﴿
A=1‐\(\frac{1}{1024}\)
= \(\frac{1023}{1024}\)
vậy A=\(\frac{1023}{1024}\)
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=-1.\)
b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=6.\frac{5}{4}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}.\)
c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)
\(=\frac{6}{7}.\)
d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)
\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)
\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)
\(=\frac{107}{100}+\frac{1}{5}\)
\(=\frac{127}{100}.\)
Chúc bạn học tốt!
a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{-59}{45}\)
b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)
\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)
\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)
\(\Rightarrow\frac{31}{4}\)
c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)
\(\Rightarrow\frac{6}{7}\)
d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)
\(\Rightarrow\frac{93}{100}\)
ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)
\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
đặt D=-(1/2+1/4+1/8+....+1/1024)
D=-(1/2+1/2^2+....+1/2^10)
đặt A=1/2+...+1/2^10
2A = 1+1/2+...+1/2^9
2A-A=(1+1/2+...+1/2^9)-(1/2+...+1/2^10)
A=1-1/2^10
A=2^10-1/2^10
D=-2^10-1/2^10