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đề bài là vậy :\(\frac{1}{7}x-x=\frac{\left(-3\right)^2}{5}\)
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`@` `\text {Ans}`
`\downarrow`
`3^4*5^2 - 128*2^3 + 1^17`
`= 9^2*5^2 - 2^7*2^3 + 1`
`= (9*5)^2 - 2^10+1`
`= 45^5-2^10 + 1`
`= 2025 - 1024 + 1`
`= 2025 - 1023`
`= 1002`
\(24-4^2\div4\cdot2+3\)
`= 24 - 4*2 + 3`
`= 24 - 8 + 3`
`= 24 - 5`
`= 19`
`@` `\text {Kaizuu lv u}.`
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a, 24.(82+18)-100
=(24.100).100
=2400-100
=2300
xl nha nik chỉ làm đc phần a thôi
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\(P=1+5+5^2+..........+5^{2016}\)
\(\Leftrightarrow5P=5+5^2+5^3+............+5^{2017}\)
\(\Leftrightarrow5P-P=\left(5+5^2+..........+5^{2017}\right)-\left(1+5+..........+5^{2016}\right)\)
\(\Leftrightarrow4P=5^{2017}-1\)
\(\Leftrightarrow P=\dfrac{5^{2017}-1}{4}\)
Mà \(Q=\dfrac{5^{2017}}{4}\)
\(\Leftrightarrow Q-P=\dfrac{5^{2017}}{4}-\dfrac{5^{2017}-1}{4}=\dfrac{5^{2017}-5^{2017}-1}{4}=-\dfrac{1}{4}\)
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\(D=4\times5^{100}\times\left[\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)^2\left(\frac{1}{5}\right)^3...\left(\frac{1}{5}\right)^{100}\right]+1\)
\(=4\times5^{100}\times\frac{1}{5\times5^2\times...\times5^{100}}+1\)
\(=\frac{4}{5\times5^2\times...\times5^{99}}+1\)
\(=\frac{4}{5^{99\left(1+99\right):2}}+1=\frac{4}{5^{4950}}+1\)