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<=> (32+1 ).33(x-2) =317+315
<=> (32+1 ).33(x-2) =315.(32+1)
<=> 33(x-2) = 315
<=> 3(x-2) = 15
<=> x-2 = 5
<=> x= 7
\(10.3^{3.\left(x-2\right)}=\left(3^{117}+3^{115}\right):3^{100}\)
\(10.3^{3x-9}=3^{117}:3^{100}+3^{115}:3^{100}\)
\(10.3^{3x}:3^9=3^{17}+3^{15}\)
\(10.9^x:19683=143489070\)
...................................................
10x33(x-2)=(3117+3115):3100
10x33(x-2)=3117:3100+3115:3100
10x33(x-2)=3117-100+3115-100
10x33(x-2)=317+315
10x33(x-2)=129 140 163+14 348 907
10x33(x-2)=143 489 070
33(x-2)=143 489 070:10
33(x-2)=14 348 907
33(x-2)=315
=>3(x-2)=15
x-2=15:3
x-2=5
x=5+2
x=7
Lời giải:
$10.3^x=3^{117}+3^{115}=3^{115}(3^2+1)=10.3^{115}$
$\Rightarrow 3^x=3^{115}$
$\Rightarrow x=115$
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`
a)Đặt \(A=3-3^2+3^3-3^4+...+3^{95}-3^{96}\)
\(3A=3^2-3^3+3^4-3^5+...+3^{96}-3^{97}\)
\(3A+A=\left(3^2-3^3+3^4-3^5+...+3^{96}-3^{97}\right)+\left(3-3^2+3^3-3^4+...+3^{95}-3^{96}\right)\)
\(4A=-3^{97}+3\)
\(A=\frac{-3^{97}+3}{4}\)
b)tương tự như câu a
c)\(\left(100-1^2\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-99^2\right)\)
\(=\left(10^2-1^2\right)\left(10^2-2^2\right)\left(10^2-3^2\right)....\left(10^2-10^2\right)...\left(10^2-99^2\right)\)
\(=\left(10^2-1^2\right)\left(10^2-2^2\right)\left(10^2-3^2\right)...0...\left(10^2-99^2\right)\)
=0