Question

\(\dfrac{x^3-4x}{10-5x}=\dfrac{-x^2-2x}{5}\)

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

\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

\(\dfrac{x^3+8}{x^2-2x+4}=x+2\)

Answer 1

a: \(\dfrac{x^3-4x}{10-5x}\)

\(=\dfrac{x\left(x-2\right)\left(x+2\right)}{5\left(2-x\right)}=\dfrac{-x^2-2x}{5}\)

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

c: \(\left(x^2-3x+2\right)\left(x+1\right)\)

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

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

=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

d:\(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)