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\(4S=4+4^2+4^3+...+4^{2019}\)
=>3S=4^2019-1
hay \(S=\dfrac{4^{2019}-1}{3}\)
a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)
\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)
=>3x=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 \(x\in\left\{1;2\right\}\)
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) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)
1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)
2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
a: \(-3xy^2+x^2y^2-5x^2y\)
\(=xy\left(-3y+xy-5x\right)\)
c: \(y^2+xy+y=y\left(y+x+1\right)\)
\(A=1+2+2^2+.....+2^{2018}\)
\(\Leftrightarrow2A=2+2^2+....+2^{2018}+2^{2019}\)
\(\Leftrightarrow2A-A=\left(2+2^2+....+2^{2019}\right)-\left(1+2+2^2+....+2^{2018}\right)\)
\(\Leftrightarrow A=2^{2019}-1< 2^{2019}\)
Vậy \(A< 2^{2019}\)