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a=5k+1
Cho \(0\le5k+1\le100\)
=> \(0\le k\le19\)
Có 20 giá trị của k => A có 20 phần tử
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5-\left(3x+5\right)=18-\left(-2x+3\right)\)
\(5-3x-5=18+2x-3\)
\(-3x-2x=18-3-5+5\)
\(-5x=15\)
\(x=-3\)
\(5-\left(3x+5\right)=18-\left(-2x+3\right)\)
\(\Leftrightarrow5-3x-5=18+2x-3\)
\(\Leftrightarrow-3x=15+2x\)
\(\Leftrightarrow-2x-3x=15\)
\(\Leftrightarrow-5x=15\)
\(\Leftrightarrow x=-3\)
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5 . 3 - 5 . x - 2 . x - 2 . 4 = -4
15 - 5 . x - 2 .x -8 =-4
15 - 5 .x - 2. x = -4 + 8
15 - 5 . x -2 . x = 4
5 . x - 2 .x = 15 - 4
x.( 5 - 2) =9
x. 3 9
x = 9 /3
x =3
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3+x=-2+3
3+x=1
x=1-3
x=1+(-3)
x=-2
Vậy x=-2
Nhớ tick cho mk nhé Hiền Lê hihiih
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b: \(B=5+5^2+5^3+5^4+...+5^{99}+5^{100}\)
\(=5\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=6\cdot\left(5+...+5^{99}\right)⋮6\)
a) \(A=1+3+3^2+...+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(=40+3^4.40+...+3^{96}.40=40\left(1+3^4+...+3^{96}\right)⋮40\)
![](https://rs.olm.vn/images/avt/0.png?1311)
192.(54+64)-92.118
192.118-92.118
(192-92).118
100.118
=11800
TÍCH CHO MK NHA
bn lấy máy tính ra mà tính