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\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\frac{1}{486}\)
\(3\times A=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\)
\(3\times A-A=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\frac{1}{486}\right)\)
\(2\times A=\frac{3}{2}-\frac{1}{486}\)
\(A=\frac{182}{243}\)
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B)A*2=(1/2+1/4+....+1/256)*2
=1+1/2+1/4+....+1/128)
A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)
=1-1/256
=255/256
a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)
\(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)
Lấy \(A-\frac{1}{3}\times A\)theo vế ta có :
\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)
\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)
\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)
\(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)
\(\Rightarrow A=\frac{605}{162}\)
Vậy \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)
b) Đặt B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)
Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có :
\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)
\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)
\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)
\(\Rightarrow B=\frac{255}{256}\)
Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)
Đặt tổng trên là A
\(3xA=\dfrac{15}{2}+\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{18}+\dfrac{5}{54}+\dfrac{5}{162}\)
\(2xA=3xA-A=\dfrac{15}{2}-\dfrac{5}{468}\)
Từ đó suy ra A
\(\frac{1\cdot3\cdot9+2\cdot6\cdot18+3\cdot9\cdot27}{1\cdot5\cdot18+2\cdot10\cdot36+3\cdot15\cdot54}\)
\(=\frac{1\cdot3\cdot9+2\left(1\cdot3\cdot9\right)+3\left(1\cdot3\cdot9\right)}{1\cdot5\cdot18+2\left(1\cdot5\cdot18\right)+3\left(1\cdot5\cdot18\right)}\)
\(=\frac{\left(1\cdot3\cdot9\right)\left(1+2+3\right)}{\left(1\cdot5\cdot18\right)\left(1+2+3\right)}\)
\(=\frac{3}{10}\)
A) bạn xem lại đề ạ
B) 1/2 + 1/6 + 1/ 12 + 1/ 20 + ...+ 1/ 9900
=1/2+1/6+1/12+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100
=99/100
C) Biến đổi tử số và mẫu số ta có
- Tử số: 20,2 x 5,1 - 30,3 x 3,4 + 14,58
= 103,02 - 103,02 + 14,58
= 14,58
- Mẫu số: 14,58 x 460 + 7,29 x 540 x 2
= 14,58 x 460 + 14,58 x 540
= 14,58 x (460 + 540)
= 14,58 x 1000
= 14580
Thay vào ta có: = 14,58 : 14580
= 0,001
Vậy 20.2*5.1-30.3*3.4+14.56/ 14.58*460+7.29 *540*2 = 0,001.
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\)
\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)
Gọi \(B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow B-\frac{1}{3}B=1-\frac{1}{243}\)
\(\Rightarrow\frac{2}{3}B=\frac{242}{243}\)
\(\Rightarrow B=\frac{121}{81}\)
Suy ra \(A=\frac{1}{2}B=\frac{1}{2}.\frac{121}{81}=\frac{121}{162}\)