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`4(x-5)-2^3=2^{4}.3`
`=>4(x-5)-8=16.3=48`
`=>4(x-5)=48+8=56`
`=>x-5=56:4=14`
`=>x=19`
Vậy `x=19`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5(x+7)-10=23.5
5(x+7)-10=8.5
5(x+7)-10=45
5(x+7)=55 (45+10)
x+7=\(\dfrac{55}{5}\)=11 =>x=11-7=3
Ta có: \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Leftrightarrow5\left(x+7\right)=50\)
\(\Leftrightarrow x+7=10\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
5(x+7)-10=23.5
5(x+7)-10=8.5
5(x+7)-10=45
5(x+7)=55
x+7=\(\dfrac{55}{5}\)=11 =>x=11-7=3
Ta có: \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Leftrightarrow5\left(x+7\right)=50\)
\(\Leftrightarrow x+7=10\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
24 - 2 . (15 - x) = 10
16 - 2 . (15 - x) = 10
2 . (15 - x) = 16 - 10
2 . (15 - x) = 6
15 - x = 6 : 2
15 - x = 3
x = 15 - 3
x = 12
Chúc bạn học tốt!! ^^
![](https://rs.olm.vn/images/avt/0.png?1311)
24 - 2(15 - x) = 10
16 - 2(15 - x) = 10
2(15 - x) = 16 -10
2(15 - x) = 6
15 - x = 6 : 2
15 - x = 3
x = 15 - 3
x = 12
Vậy x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 21.52.17 = 2.25.17 = 50.17 = 850
b, 22 + 23 + 24 = 4 + 8 + 16 = 28
c, 25.3 + 24:8 + 50: 52
= 32.3 + 16:8 + 50:25
=96 + 2 + 2
= 100
d, 112 - 102 - 32
= 121 - 100 - 9
= 21 - 9
= 12
e, 13 + 23 + 33 + 43 + 53
= ( 1+ 2+3+4+5)2
= 152
= 225
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰
= 2 + 2².(1 + 2 + 2²) + 2⁵.(1 + 2 + 2²) + ... + 2⁹⁸.(1 + 2 + 2²)
= 2 + 7.2² + 7.2⁵ + ... + 7.2⁹⁸)
= 2 + 7.(2² + 2⁵ + ... + 2⁹⁸)
Vậy số dư khi chia A cho 7 là 2
\(A=2+2^2+2^3+2^4+2^5+...+2^{100}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)+2^{100}\)
\(=2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{97}\left(1+2+4\right)+2^{100}\)
\(=7\left(2+2^4+...+2^{97}\right)+2^{100}\)
\(Vì7⋮7=>7\left(2+2^4+..+2^{97}\right)⋮7\)
Ta có:
\(2^3\equiv1\left(mod7\right)\)
\(2^{3.33}\equiv1^{33}\left(mod7\right)\equiv1\left(mod7\right)\)
\(2^{3.33}=2^{99}=>2^{100}=2^{99}.2\equiv1.2\left(mod7\right)\equiv2\left(mod7\right)\)
\(=>2^{100}\) chia \(7\) dư \(2\) mà \(7\left(2+2^4+...+2^{97}\right)⋮7\)
\(=>A\) chia \(7\) dư \(2\)
4(x - 5) - 23 = 24 . 3
4(x - 5) - 8 = 48
4(x - 5) = 48 + 8
4(x - 5) = 56
x - 5 = 56 : 4
x - 5 = 14
x = 14 + 5
x = 19
Vậy x = 19
4( x - 5 ) - 2^3 = 2^4 . 3
4( x - 5 ) - 8 = 16 . 3
4( x - 5 ) - 8 = 48
4( x - 5 ) = 48 + 8
4( x - 5 ) = 56
x - 5 = 56 : 4
x - 5 = 14
x = 14 + 5
x = 19