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\(5S=5+1+\frac{1}{5^2}+...+\)\(\frac{1}{5^{2019}}\)
\(5S-S=4S=5-\frac{1}{5^{2019}}\)
\(\Rightarrow S=\frac{\frac{5^{2020}}{5^{2019}}}{4}\)
\(S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\)
\(5S=5\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)
\(5S=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)
\(5S-S=4S\)
\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)
\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-1-\frac{1}{5}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2020}}\)
\(=5-\frac{1}{5^{2020}}\)
\(4S=5-\frac{1}{5^{2020}}\Rightarrow S=\frac{5-\frac{1}{5^{2020}}}{4}\)
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a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
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a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)
b: =78(0,65+0,35)+2020(2,2-2,2)
=78*1=78
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Gọi 1+5-1+5-2+5-3+...+5-2020=A
Ta có 5A=5(1+5-1+5-2+5-3+...+5-2020)
=51+50+5-1+5-2+5-3+...+5-2019
⇒5A-A=(51+50+5-1+5-2+5-3+...+5-2019)-(1+5-1+5-2+5-3+...+5-2020)
\(\Rightarrow\)4A=51-5-2020
\(\Rightarrow\)A=\(\frac{5^1-5^{-2020}}{4}\)
\(\frac{5^1-5^{-2020}}{4}\)\(\frac{5^1-5^{-2020}}{4}\)Vậy A=\(\frac{5^1-5^{-2020}}{4}\)\(\frac{5^1-5^{-2020}}{4}\)
\(A=1+5^{-1}+5^{-2}+5^{-2}+...+5^{-100}\)
\(\Rightarrow5A=5+1+5^{-1}+5^{-2}+...+5^{-99}\)
\(\Rightarrow5A-A=5+1+5^{-1}+5^{-2}+...+5^{-99}-1-5^{-1}-5^{-2}-5^{-3}-...-5^{-100}\)
\(\Leftrightarrow4A=5-5^{-100}\)
\(\Rightarrow A=\frac{5-5^{-100}}{4}\)
Chúc bạn học tốt@@
A= \(\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{6}+\dfrac{7}{8}+...+\dfrac{2019}{2020}\)
tính A.
help mik với
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