Question

1) rút gọn 

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

b) \(\dfrac{x^3+3x^2-2}{x^3+3x+4}=\)

Answer 1

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

Answer 2

\(a,\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\dfrac{\left(x^2-y^2\right)+\left(3x-3y\right)}{x^2-y^2}\\ =\dfrac{\left(x-y\right)\left(x+y\right)+3\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y+3\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+3}{x+y}\)