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\(\frac{2^{500}-5.2^{495}}{2^{501}-10.2^{495}}\)=\(\frac{2^{495}.\left(2^5-5\right)}{2^{495}.\left(2^6-10\right)}=\frac{2^5-5}{2^6-10}=\frac{32-5}{64-10}=\frac{27}{54}=\frac{1}{2}\)
\(A=\frac{2^{500}-5.2^{495}}{2^{501}-10.2^{495}}=\frac{2^{495}.2^5-5.2^{495}}{2^{495}.2^6-10.2^{495}}=\frac{2^{495}\left(2^5-5\right)}{2^{495}\left(2^6-10\right)}=\frac{2^5-5}{2^6-10}=\frac{1}{2}\)
a) \(5^5\)
b) \(5^7\)
c) \(5^{19}\)
d) \(10^3\)
e) \(10.10^2.10^4.10^4=10^{11}\)
`@` `\text {Ans}`
`\downarrow`
`(7x - 11)^2 = 2^5 * 5^2 + 100?`
`=> (7x - 11)^2 = 32*25 + 100`
`=> (7x - 11)^2 = 800 + 100`
`=> (7x - 11)^2 = 900`
`=> (7x - 11)^2 = (+-30)^2`
`=>`\(\left[{}\begin{matrix}7x-11=30\\7x-11=-30\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}7x=41\\7x=-19\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{41}{7}\\x=-\dfrac{19}{7}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{\dfrac{41}{7};-\dfrac{19}{7}\right\}\)
2A=\(1+\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
2A-A=\(1-\frac{1}{2^{100}}\)
A=\(\frac{2^{100}-1}{2^{100}}\)
a) \(7.7.8.8.8=7^2.8^3\)
b) \(10.2.5.5.2.100=10.2.5.5.2.10.10=10^3.2^2.5^2\)
a, \(7^2\times8^3\)
\(b.10^3\times2^2\times5^2\)
Ủng hộ tớ nhé bạn !♥