Bài 2:
\(a,=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\\ =\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\\ b,=\dfrac{x^3-xy^2-x^2y+xy^2-x^3}{\left(x-y\right)\left(x+y\right)}=\dfrac{-x^2y}{\left(x-y\right)\left(x+y\right)}\)
Bài 3:
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\\ =\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}\\ =\dfrac{y}{x-y}-\dfrac{x}{x-y}=\dfrac{y-x}{x-y}=-1\)
Bài 2:
a: \(=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\)
