\(A=1+3+3^2+...+3^{59}\\ =\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+\left(3^{54}+3^{55}+3^{56}+3^{57}+3^{58}+3^{59}\right)\\ =1\left(1+3+3^2+3^3+3^4+3^5\right)+3^6\left(1+3+3^2+3^3+3^4+3^5\right)+...+3^{54}\left(1+3+3^2+3^3+3^4+3^5\right)\\ =\left(1+3+3^2+3^3+3^4+3^5\right)\left(1+3^6+...+3^{54}\right)\\ =364\left(1+3^6+...+3^{54}\right)\\ =4\cdot13\cdot7\left(1+3^6+...+3^{54}\right)\text{ chia hết cho 4 và 13}\)

a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)

b) \(\left(4x-1\right)^2=16x^2-8x+1\)

c) \(\left(x+2\right)^2=x^2+4x+4\)

d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)

e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)

f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)

g) \(\left(x+1\right)^3=x^3+3x^2+3x+1\)

h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)

k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)

^{ai Giúp mình với}

a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)

\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)

=>3^x=3⁷

hay x=7

b: \(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)

\(\Leftrightarrow4^{x+1}\left(4^2-3\right)=13\cdot4^{11}\)

=>x+1=11

hay x=10

d: \(\left(x-1\right)^{13}=\left(x-1\right)^{12}\)

\(\Leftrightarrow\left(x-1\right)^{12}\left(x-2\right)=0\)

hay \(x\in\left\{1;2\right\}\)

Câu hỏi lớp 6 mà mk ghi lộn

\(4S=4+4^2+4^3+...+4^{2019}\)

=>3S=4^2019-1

hay \(S=\dfrac{4^{2019}-1}{3}\)

3C=3^4+3^5+...+3^2019

=>2C=3^2019-3

=>\(C=\dfrac{3^{2019}-3}{2}\)

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

\(3S=3+3^2+...+3^{2009}+3^{2010}\)

\(3S-S=3^{2010}-1\)

\(S=\frac{3^{2010}-1}{2}\)

\(A=1+2+2^2+.....+2^{2018}\)

\(\Leftrightarrow2A=2+2^2+....+2^{2018}+2^{2019}\)

\(\Leftrightarrow2A-A=\left(2+2^2+....+2^{2019}\right)-\left(1+2+2^2+....+2^{2018}\right)\)

\(\Leftrightarrow A=2^{2019}-1< 2^{2019}\)

Vậy \(A< 2^{2019}\)