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A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
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HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
16 tháng 9 2023
a) `1/9-0,3. 5/9+1/3`
`=1/9-3/10 . 5/9+1/3`
`=1/9-15/90+1/3`
`=1/9-1/6+1/3`
`=2/18-3/18+6/18`
`=5/18`
b) `(-2/3)^2+1/6-(-0,5)^3`
`=4/9+1/6-(-0,125)`
`=4/9+1/6+0,125`
`=4/9+1/6+1/8`
`=32/72+12/72+9/72`
`=53/72`
2 tháng 6 2017
\(A=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{100}\right)\)
\(A=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{101}{100}\)
\(A=\frac{101}{2}\)
VT
30 tháng 4 2017
A = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{101}{100}\)
A = \(\frac{101}{2}\)
HQ
Hà Quang Minh
Giáo viên
2 tháng 11 2023
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$
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
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)
\(=\frac{49}{34}\)
AM
2 tháng 6 2015
Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}+\frac{1}{3^9}\)
=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>3A-A=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\)\(\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}+\frac{1}{3^9}\right)\)
=>2A=\(1-\frac{1}{3^9}\)
=>A=\(\frac{9841}{19683}\)
WR
2 tháng 6 2015
đặt biểu thức trên là A
ta có
3A=3(1/3+1/3^2+1/3^3+1/3^4+....+1/3^9)
3A=1+1/3+1/3^2+...+1/3^8
3A-A=1+1/3+1/3^2+...+1/3^8-(1/3+1/3^2+1/3^3+..+1/3^9)
2A=1-1/3^9
2A=3^9-1/3^9
A=3^9-1/3^9.2
vậy A=3^9-1/3^9.2
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=1-\frac{1}{3^8}\)
\(2A=1-\frac{1}{3^8}\)
\(A=\frac{1-\frac{1}{3^8}}{2}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^8}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)
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