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ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)
\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
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
nhầm dấu, ngoặc đầu tiên là dấu trừ chứ không phải dấu cộng
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
\(\text{Ta có: }\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{8}\right)-......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Giả sử A = -B
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=2-\frac{1}{1024}\)
\(B=\frac{2047}{1024}\)
=> \(A=-\frac{2047}{1024}\)
\(A=\frac{2047}{1024}\)
k mk nha
mkt ick lại cho!