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#)Giải :
Bài 1 :
\(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\right)-\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-A-\frac{1}{1280}\)
\(\Rightarrow A\times2+A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{511}{1280}\)
#)Giải :
Bài 2 :
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{59049}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\right)\)
\(2B=1-\frac{1}{3^{10}}\)
\(B=\frac{1-\frac{1}{3^{10}}}{2}\)
A=1/3+1/9+1/27+1/81+1/243+1/729
3A=1+1/3+1/9+1/27+1/81+1/243
3A-A=(1+1/3+1/9+1/27+1/81+1/243)-(1/3+1/9+1/27+1/81+1/243+1/729)
3A-A=1-1/3+1/3-1/9+1/9-1/27+1/27-1/81+1/81-1/243+1/243-1/729)
2A=1-1/729
2A=728/729
A=728/729/2
A=364/729
\(=1\frac{364}{729}\)\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}=1+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}=1\frac{ }{ }\)
1. Số vở thầy phát là :
38 - 6 = 32 ( quyển )
Số vở lúc đầu thầy có là :
32 : 2 x 5 = 80 ( quyển )
Đáp số : ...
2. Tự phân tích từng số :
Số 1/3 = 19683 / 59049
Lấy 59049 chia cho mẫu số số cần chuyển , ra tử số cần chuyển , lấy mẫu là 59049
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=\(1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
=\(\frac{3^6}{3^6}+\frac{3^5}{3^6}+\frac{3^4}{3^6}+\frac{3^3}{3^6}+\frac{3^2}{3^6}+\frac{3^1}{3^6}+\frac{3^0}{3^6}\)
=\(\frac{3^6+3^5+3^4+3^3+3^2+3+1}{3^6}\)
=\(\frac{729+243+81+27+9+3}{729}\)
=\(\frac{1093}{729}\)
nha.
1 + 1/3 + 1/9+1/27+1/81+1/243+1/729
=1+1-1/3+1/3-1/9+1/9-1/27-1/27-1/81+1/81-1/243
= 2 - 1/243
=485/243
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{729}{729}+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{729+243+81+27+9+3+1}{729}\)
\(=\frac{1087}{729}\)
Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3A-A=3-\frac{1}{729}\)nên \(2A=3-\frac{1}{729}\)
kHI ĐÓ \(A=\left(3-\frac{1}{729}\right):2=\frac{3}{2}-\frac{1}{1458}=\frac{2197}{1458}-\frac{1}{1458}=\frac{2196}{1458}\)
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times A-A=3-\frac{1}{729}=\frac{2186}{729}\)
\(2\times A=\frac{2186}{729}=>A=\frac{1093}{729}\)
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
- ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049
A=1/3+1/9+1/27+1/81+......+1/95049
Ax3=3x(1/3+1/9+1/27+1/81+.........+1/95049)
Ax3=1+1/3+1/9+1/27+1/81+..........+1/19683
Ax3-A=1+1/3+1/9+1/27+1/81+............+1/19683
- (1+1/3+1/9+1/27+1/81+........+1/59049)
=1-1/59049
2xA=59048/59049
A=59048/59049:2
A=29524/59049