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Lời giải:
$A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2023}}$
$2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2022}}$
$2A-A=2-\frac{1}{2^{2023}}$
$A=2-\frac{1}{2^{2023}}$
q=1/3; u1=2/3
\(S_{100}=\dfrac{\dfrac{2}{3}\cdot\left(\dfrac{1}{3^{100}}-1\right)}{\dfrac{1}{3}-1}=-\dfrac{1}{3^{100}}+1=\dfrac{-1+3^{100}}{3^{100}}\)
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
\(\dfrac{2^3.3^4}{2^3.3^2.5}=\dfrac{1.3^2}{1.1.5}=\dfrac{9}{5}\)
a: \(=\dfrac{13\cdot\left(9-2\right)}{13}=7\)
b: \(=\dfrac{14\left(3-8\right)}{7\left(1+3\cdot3\right)}=2\cdot\dfrac{-5}{10}=-1\)
c: \(=\dfrac{54-72}{36}=\dfrac{-18}{36}=-\dfrac{1}{2}\)
d: \(=\dfrac{5^3}{10^2\cdot5}=\dfrac{5^2}{100}=\dfrac{1}{4}\)
`(4^2. 5.11)/(44.20)`
`=(4.11.4.5)/(4.11.4.5)`
`=1`
`(13.15.16)/(18.65.7)`
`=(13.15.16)/(2.3.3.13.5.7)`
`=8/21`
`(7.2.8.5^2)/(14.2.5)`
`=(14.2.4.5.5)/(14.2.5)`
`=4.5`
`=20`
`(2^3. 3^3. 5)/(3.2^3. 5^3)`
`=(2^3. 3.5.3^2)/(2^3. 3.5.5^2)`
`=(3^2)/(5^2)`
`=9/25`
**Quy đồng:
`(4^2. 5.11)/(44.20)=1=525/525`
`(13.15.16)/(18.65.7)=8/21=200/525`
`(7.2.8.5^2)/(14.2.5)=20=840/525`
`(2^3. 3^3. 5)/(3.2^3. 5^3)=9/25=189/525`
A=1/3-1/3^2+...-1/3^20
=>3A=1-1/3+...-1/3^19
=>4A=1-1/3^20
=>\(A=\dfrac{3^{20}-1}{3^{20}\cdot4}\)
\(\dfrac{4\cdot5\cdot36}{35\cdot9\cdot2}=\dfrac{2^2\cdot5\cdot2^2\cdot3^2}{5\cdot7\cdot3^2\cdot2}=\dfrac{2^3}{7}=\dfrac{8}{7}\left(C\right)\)
\(\dfrac{2^4\cdot125}{2^7\cdot50}=\dfrac{2^4\cdot5^3}{2^8\cdot5^2}=\dfrac{5}{2^4}=\dfrac{5}{16}\)