Question

Answer 1

a: \(\dfrac{A}{x^3-8}=\dfrac{x-1}{x^2+2x+4}\)

\(\Leftrightarrow A=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)\left(x-1\right)}{x^2+2x+4}=x^2-3x+2\)

b: \(\Leftrightarrow\dfrac{1}{B}=\dfrac{1}{\left(x-1\right)\left(x-3\right)}:\dfrac{2x-1}{x-3}=\dfrac{1}{\left(x-1\right)\left(2x-1\right)}\)

=>B=(x-1)(2x-1)=2x^2-3x+1