Question

chứng minh rằng: \(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\dfrac{1}{x-y}\)

Answer 1

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

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

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

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

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

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

\(=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)