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XO
10 tháng 3 2020
Ta có : \(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+....+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)(1)
=> 32.A = \(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^8}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\)(2)
Lấy (2) cộng (1) theo vế ta có :
32.A + A = \(\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}\right)+\left(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\right)\)
10A = \(1-\frac{1}{3^{100}}\)
=> A = \(\left(1-\frac{1}{3^{100}}\right):10=\frac{1}{10}-\frac{1}{3^{100}.10}=0,1-\frac{1}{3^{100}.10}< 0,1\)
=> A < 0,1 (ĐPCM)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)
Chú ý:
Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
TN
21 tháng 8 2015
\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}+\frac{1}{3^9}\)
\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
\(3A-A=\frac{1}{3}-\frac{1}{3^9}\)
\(2A=\frac{1}{3}.\left(1-\frac{1}{3^8}\right)\)
\(A=\frac{1}{6}.\left(1-\frac{1}{3^8}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n-1}}+\frac{1}{2^n}\)
\(\frac{1}{2}B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^n}+\frac{1}{2^{n+1}}\)
\(B-\frac{1}{2}B=1-\frac{1}{2^{n+1}}\)
\(\frac{1}{2}B=1-\frac{1}{2^{n+1}}\)
\(B=2-\frac{2}{2^n.2}=2-\frac{1}{2^n}\)
LC
1
M
11 tháng 9 2019
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+...+\frac{1}{3^8}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^8}\)
\(\Rightarrow A=\left(1-\frac{1}{3^8}\right):2\)
\(\Rightarrow A=\frac{3280}{6561}\)
XO
4 tháng 7 2019
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)
= \(\frac{15}{8}+\frac{5}{8}\)
= \(\frac{5}{2}\)
4 tháng 7 2019
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)
\(=\frac{49}{34}\)
24 tháng 6 2018
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
24 tháng 6 2018
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
Ta có:
3A=1+\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^7}\)
=>3A-A=1+\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^7}\)-(\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^8}\))
<=>2A=1-\(\frac{1}{3^8}\)
=> A=\(\frac{3^8-1}{2.3^8}\)=\(\frac{3280}{6561}\)
3×A= 1+ 1/3+1/32+ 1/33+...+1/37
3×A-A= (1+1/3+1/32+...+1/37) -( 1/3+1/32+....+1/38)
2×A= 1-1/38
A= (1- 1/38):2
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