\(\dfrac{x-x^2}{5x^2-5}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{5\left(x^2-1\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{-5\left(1-x^2\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{-5\left(1-x\right)\left(1+x\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x}{-5\left(1+x\right)}=\dfrac{x}{A}\\ \Leftrightarrow A=-5\left(1+x\right)\\ \Leftrightarrow A=-5x-5\)
<=>\(\dfrac{x-x^2}{5x^2-5}\)= \(\dfrac{x\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}\)= \(\dfrac{x}{5\left(x+1\right)}\)
Vậy đa thức A = 5(x+1)
