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\(3+3^2+.....+3^{99}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(=39+3^3\left(3+3^2+3^3\right)+........+3^{96}\left(3+3^2+3^3\right)\)
\(=39+3^3\cdot39+...+3^{96}\cdot39\)
\(=39\left(1+3^3+....+3^{96}\right)\)
Vì \(39⋮13\Rightarrow39\in B\left(13\right)\)
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b)
=>2x+1=3/5
2x=3/5-1
2x=-2/5
x=-2/5:2
x=-1/5
hoặc:
2x+1=-3/5
2x=-3/5-1
2x=-8/5
x=-8/5:2
x=-4/5
=>x=-1/5 hoặc x=-4/5
\(a.x^{2017}=x\)
Vì 0^2017 = 0
1^2017 = 1
=> x = 0 hoặc x = 1
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3\(^{2x}\) = 81
=> 3\(^{2x}\) = 3\(^4\)
=> 2x = 4
=> x = 2
Vậy....
Ta có: 3^2x-1=80
<=> 3^2x=81
<=> 3^2x=3^4
<=> 2x=4
<=> x=2
Vậy x=2
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3600/[(5x+335)/x]=50
(5x+355)/x=3600/50
5x+355=72x
72x-5x=355
67x=355
x=355/67
ta có 3600 : [(5x+335) : x] = 50
<=>3600*\(\frac{x}{5x+335}\)= 50
áp dụng quy tắc chéo
<=>3600*x =50* (5x+335)
<=>3600x = 250x +16750
<=>3600x-250x =16750
<=>3350x = 16750
<=> x =5
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P = (10.10².10³.10⁴...10⁹) : (10⁵.10¹⁰.10²⁵)
= 10¹⁺²⁺³⁺⁴⁺⁵⁺⁶⁺⁷⁺⁸⁺⁹ 10⁵⁺¹⁰⁺²⁵
= 10⁴⁵ : 10⁴⁰
= 10⁴⁵⁻⁴⁰
= 10⁵
= 100000
\(P=\left(10.10^2.10^3.10^4.....10^9\right):\left(10^5.10^{10}.10^{25}\right)\)
\(P=10^{45}:10^{40}\)
\(P=10^5\)
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ta co:
6n +11 chia het 2n + 1
=> 6n +11 - 3( 2n +1 ) chia het 2n +1
=> 6n +11 - 6n - 3 chia het 2n +1
=> 8 chia het 2n +1
=> 2n +1 thuoc uoc cua 8
=> 2n +1 thuoc {.......} ban tu liet ke nhe!
=> 2n thuoc { .........} ban tu kiet ke
=> n thuoc {.......} ban tu kiet ke
=> n= -1
tick nha!
<=>3(n+1)+10 chia hết 2n+1
=>30 chia hết 2n+1
=>2n+1\(\in\)U(30)={....} bạn tự liệt kê
=>n\(\in\){....} lấy U(30) chia cho 2 rồi -1
=7x32+2017-a -2x32017-a
=32017-ax(7x9-2)
=32017-ax61 là snt khi và chỉ khi 32017-a=1 hay a=2017
chúc bạn học tốt
HYC-23/1/2022