\(3^{x-1}.7+3^{x-1}.2=9\\ 3^{x-1}.\left(7+2\right)=9\\ 3^{x-1}.9=9\\ 3^{x-1}=\dfrac{9}{9}=1\\ Mà:3^0=1\\ Nên:x-1=0\\ Vậy:x=0+1=1\\ ---\\ P=2+2^2+2^3+...+2^{65}+2^{66}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{64}+2^{65}+2^{66}\right)\\ =2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{64}\left(1+2+2^2\right)\\ =2.7+2^4.7+...+2^{64}.7\\ =\left(2+2^4+....+2^{64}\right).7⋮7\left(đpcm\right)\)

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\(3^{x-1}.7+3^{x-1}.2=9\)
\(3^{x-1}.\left(7+2\right)=9\)
\(3^{x-1}.9=9\)
\(3^{x-1}=9:9\)
\(3^{x-1}=1\)
⇔\(3^{x-1}=3^0\)
⇒\(x-1=0\)
\(x=0+1\)
\(x=1\)
Vậy \(x=1\)

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\(2+2^2+2^3+...+2^{65}+2^{66}\)
Vì \(2+2^2+2^3=14\) mà \(14\)⋮\(7\)
⇒Ta nhóm 3 số với nhau
Ta có:
\(\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{64}+2^{65}+2^{66}\right)\)
\(\left(2+2^2+2^3\right)+2^3.\left(2+2^2+2^3\right)+...+2^{63}.\left(2+2^2+2^3\right)\)
\(14.1+14.2^3+...+14.2^{63}\)
\(14.\left(1+2^3+...+2^{63}\right)\)
Do \(14\)⋮\(7\) nên \(P=14.\left(2+2^3+...+2^{63}\right)\)⋮\(7\)

Xin tick

Ta có : 

\(A=2+2^2+2^3+2^4...2^{2010}\)\(^0\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=2.3+2^3.3+....+2^{2009}.3\)

\(=3\left(2+2^3+....+2^{2009}\right)⋮3\)

Ta có :

\(2+2^2+2^3+2^4+....+2^{2010}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+....+2^{2008}.7\)

\(=7\left(2+2^4+....+2^{2008}\right)⋮7\)

Vậy \(2^1+2^2+2^3+2^4+...+2^{2010}⋮3\) và \(7\)

a: \(=2^2\left(1+2\right)+2^4\left(1+2\right)=3\left(2^2+2^4\right)⋮3\)

b: \(=4^{20}\left(1+4\right)+4^{22}\left(1+4\right)=5\left(4^{20}+4^{22}\right)⋮5\)

c: \(A=\left(1+4+4^2\right)+...+4^{96}\left(1+4+4^2\right)\)

\(=21\left(1+...+4^{96}\right)⋮21\)

d: \(B=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{35}\left(1+7\right)\)

\(=8\left(7+7^3+...+7^{35}\right)⋮8\)

\(B=7\left(1+7+7^2\right)+...+7^{34}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{34}\right)\) chia hếtcho 3 và 19

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=14+2^3\cdot14+...+2^{117}\cdot14\)

\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=62+2^5\cdot62+...+2^{115}\cdot62\)

\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=126+126\cdot2^6+...+126\cdot2^{114}\)

\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)

\(A=2^1+2^2+2^3+...+2^{2016}\)

\(\Rightarrow A=2\left(1+2^1+2^2\right)+2^4\left(1+2^1+2^2\right)...+2^{2014}\left(1+2^1+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7...+2^{2014}.7\)

\(\Rightarrow A=7\left(2+2^4...+2^{2014}\right)⋮7\)

\(\Rightarrow dpcm\)

giúp mik với 

\(A=2^0+2^1+2^2+...+2^{59}\)

\(=2^0\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{57}\left(1+2+2^2\right)\)

\(=2^0.7+2^3.7+...+2^{57}.7\)

\(=7\left(2^0+2^3+...+2^{57}\right)⋮7\)