Question

thực hiện phép tính

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

Answer 1

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

\(=\dfrac{1}{x-y}-\left(\dfrac{-3xy}{x^3-y^3}\right)+\dfrac{x-y}{x^2+xy+y^2}\)

\(=\dfrac{1}{x-y}+\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)

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

Answer 2

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

nhân ở tử ra rồi thu gọn được:

\(\dfrac{2x^2+2y^2+2xy}{\left(x-y\right)\left(x^2+xy+y^2_{ }\right)}\)