\(a,A=\dfrac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x-y\right)^3}=-\dfrac{x^2+xy+y^2}{x^2-2xy+y^2}\)b,\(B=\dfrac{x^5+x+1}{x^3+x^2+x}=\dfrac{(x^5-x^2)+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{x^2\left(x^3-1\right)+\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}\)\(=\dfrac{x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}=\dfrac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}\)\(\dfrac{x^3-x^2+1}{x}\)
c, Sửa lại:\(x^2-4x+5\rightarrow x^2-4x-5\)
\(\dfrac{2x^2-x-3}{x^2-4x-5}=\dfrac{2x^2+2x-3x-3}{x^2+x-5x-5}=\dfrac{2x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)-5\left(x+1\right)}=\dfrac{\left(x+1\right)\left(2x-3\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{2x-3}{x-5}\)
a) \(\dfrac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)^3}=\dfrac{-x^2-xy-y^2}{\left(x-y\right)^2}\)b)
\(\dfrac{x^5+x+1}{x^3+x^2+x}=\dfrac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{x^2\left(x^3-1\right)+x^2+x+1}{x\left(x^2+x+1\right)}\)\(=\dfrac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\dfrac{x^3-x^2+1}{x}\)c) \(\dfrac{2x^2-x-3}{x^2-4x-5}=\dfrac{2x^2+2x-3x-3}{x^2+x-5x-5}=\dfrac{2x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)-5\left(x+1\right)}\)\(=\dfrac{\left(x+1\right)\left(2x-3\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{2x-3}{x-5}\)