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1)
Ta thấy 99 là số lẻ, 20y là số chẵn với mọi y
=> Để 6x + 99 = 20y thì 6x là số lẻ
=> x = 0
Thay x = 0 ta có 60 + 99 = 20y
=> 1 + 99 = 20y
=> 100 = 20y
=> y = 100 ; 20
=> y = 5
Vậy x = 0, y = 5
`Answer:`
2.
Ta có: \(M=1+3+3^2+3^3+3^4+...+3^{98}+3^{99}+3^{100}\)
\(=\left(1+3\right)+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2.\left(1+3+3^2\right)+...+3^{98}.\left(1+3+3^2\right)\)
\(=4+3^2.13+3^{98}.13\)
\(=4+13.\left(3^2+...+3^{98}\right)\)
Vậy `M` chia `13` dư `4`
Ta có: \(M=1+3+3^2+3^4+...+3^{99}+3^{100}\)
\(=1+\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1+3.\left(1+3+3^2+3^3\right)+3^5.\left(1+3+3^2+3^3\right)+...+3^{97}.\left(1+3+3^2+3^3\right)\)
\(=1+3.40+3^5.40+...+3^{97}.40\)
\(=1+40.\left(3+3^5+...+3^{97}\right)\)
Mà ta thấy \(40.\left(3+3^5+...+3^{97}\right)⋮40\)
Vậy `M` chia `40` dư `1`
![](https://rs.olm.vn/images/avt/0.png?1311)
4x2 = 1102
<=> 22x2 = 1102
<=> (2x)2 = 1102
=> 2x = 110
=> x = 55
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4.x^2=110^2\)
\(\Leftrightarrow2^2.x^2=110^2\)
\(\Rightarrow2.x=110\)
\(\Rightarrow x=110:2\)
\(\Rightarrow x=55\)
Vậy \(x=55\)
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\(3^2.\left(x+4\right)-5^2=5.2^2\)
= \(9.\left(x+4\right)-25=5.4\)
= \(9.\left(x+4\right)-25=20\)
= \(9.\left(x+4\right)=20+25\)
= \(9.\left(x+4\right)=45\)
\(\left(x+4\right)=45:9\)
\(x+4=5\)
\(x=5-4\)
\(x=1\)
32.(x+4)-52=5.22
32.(x+4)=52+(5.22)
32.(x+4)=45
x+4=45:32
x+4=5
x=5-4
x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^{x+2}2^{2x+1}=1152\)
\(\Rightarrow4^x.4^2+2^{2x}.2^1=1152\)
\(\Rightarrow4^x.4^2+4^x.2=1152\)
\(\Rightarrow4^x\left(4^2+2\right)=1152\)
\(\Rightarrow4^x=1152:\left(4^2+2\right)\)
\(\Rightarrow4^x=64\)
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3\)
\(4^{x+2}+2^{2x+1}=1152\)
\(\Rightarrow2^{2x+4}+2^{2x+1}=1152\)
\(\Rightarrow2^{2x+1}.\left(2^3+1\right)=1152\)
\(\Rightarrow2^{2x+1}.9=1152\)
\(\Rightarrow2^{2x+1}=128\)
\(\Rightarrow2^{2x+1}=2^7\)
\(\Rightarrow2x+1=7\)
\(\Rightarrow x=3\)
Vậu x = 3
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
B=3+3^2+3^3+.......+3^200
3B=3(3+3^2+3^3+.......+3^200)
3B= 3^2+3^3+.......+3^200+3^201
-
B=3+3^2+3^3+.......+3^200
2B=3^201-3
2B+3=3^201
Mà đề bài cho 2B+3=3^n
=> n=201
Vậy .........
Ta có:
B=3+3^2+3^3+.......+3^200
3B=3(3+3^2+3^3+.......+3^200)
3B= 3^2+3^3+.......+3^200+3^201
-
B=3+3^2+3^3+.......+3^200
2B=3^201-3
2B+3=3^201
Mà đề bài cho 2B+3=3^n
=> n=201
Vậy .........
2x = 4
2x = 22 (22 là 2 x 2 = 4 nên 4 bằng 22)
=> x = 2
ctrl + f5
#Inosuko
X=2 nha bạn hk tốt