a) x​​³ - 3x² + 3x - 1 = 0

<=>x³-x²-2x²-2x-x-1=0

<=>x²(x-1)-2x(x-1)+(x-1)=0

<=>(x²-2x+1)(x-1)=0

<=>(x-1)(x-1)(x-1)=0

<=>(x-1)³=0

<=>x=1

b) \(x^3-5x^2+4x-20=0\)

\(=\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

\(=x^2\left(x-5\right)+4\left(x-5\right)=0\)

\(=\left(x^2+4\right)\left(x-5\right)=0\)

\(x^2\ge0\)

\(\Rightarrow x^2+4\ge4>0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

`#3107.101107`

a)

\(5\left(x-1\right)^3=40\\\Rightarrow\left(x-1\right)^3=40\div5\\ \Rightarrow\left(x-1\right)^3=8\\ \Rightarrow\left(x-1\right)^3=2^3\\ \Rightarrow x-1=2\\ \Rightarrow x=2+1\\ \Rightarrow x=3\)

Vậy, `x = 3`

b)

\(3^{2x+1}+9^x=324?\\ \Rightarrow3^{2x}\cdot3+3^{2x}=324\\ \Rightarrow3^{2x}\cdot\left(3+1\right)=324\\ \Rightarrow3^{2x}\cdot4=324\\ \Rightarrow3^{2x}=81\\ \Rightarrow3^{2x}=3^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

c)

\(5^x-13=3\cdot2^2\\ \Rightarrow5^x-13=12\\ \Rightarrow5^x=12+13\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)

Vậy, `x = 2`

d)

\(8^x+2^{3x+1}=192\\ \Rightarrow2^{3x}+2^{3x}\cdot2=192\\ \Rightarrow2^{3x}\left(1+2\right)=192\\ \Rightarrow2^{3x}\cdot3=192\\ \Rightarrow2^{3x}=64\\ \Rightarrow2^{3x}=2^6\\ \Rightarrow3x=6\\ \Rightarrow x=2\)

Vậy, `x = 2.`

d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

\(=x^2-2x+1\)

\(=\left(x-1\right)^2\)