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=1/5+2/5+51/85+4/5
=7/5+51/85
=119/85+51/85
=170/85=2
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\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)
A = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\)
A = \(\dfrac{2}{5\times7}\)+\(\dfrac{4}{7\times11}\)+\(\dfrac{2}{11\times13}\)+\(\dfrac{4}{13\times17}\)+\(\dfrac{2}{17\times19}\)+\(\dfrac{4}{19\times23}\)+\(\dfrac{2}{23\times25}\)
A = \(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)-\(\dfrac{1}{19}\)+\(\dfrac{1}{19}\)-\(\dfrac{1}{23}\)+\(\dfrac{1}{23}\)-\(\dfrac{1}{25}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)
A = \(\dfrac{4}{25}\)
a)
x + 9,68 = 9,68, Ta có x = 0
(vì 0 cộng với số nào cũng bằng chính số đó).
b)
\(TC:\)
\(\dfrac{2}{5}=\dfrac{4}{10}\)
Vậy : x = 0
\(\dfrac{15}{10}=1,5;\dfrac{35}{100}=0,35;\dfrac{107}{100}=1,07\)
\(\dfrac{22109}{1000}=22,109;\dfrac{14}{5}=\dfrac{28}{10}=2,8;\dfrac{920}{1000}=0,92\)
\(\dfrac{138}{100}=1,38;\dfrac{2007}{10}=200,7;\dfrac{1}{1000}=0,001\)
=(1/5+3/5+4/5+2/5)
=10/5=2
\(=\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)