Ta có : x + y = 1
=> x = 1 - y
y = 1 - x , 1 - ( x + y ) = 0
Khi đó : \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{1-y}{\left(y-1\right)\left(y^2+y+1\right)}-\frac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2+x+1\right)+\left(y^2+y+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-x^2-x-1+y^2+y+1}{x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x^2-y^2\right)-\left(x-y\right)}{x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+\left(x+y\right)+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(x-y\right)\left(-x-y-1\right)}{x^2y^2+xy.1+x^2+y^2+xy+1+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(x-y\right)\left(-x-y-1\right)}{x^2y^2+\left(x+y\right)^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y-1\right)\left(x+y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y-1\right)\left(x+y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{-\left(x-y-1\right)\left(x+y\right)+2\left(x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(x-y\right)\left[-\left(x+y+1\right)+2\right]}{x^2y^2+3}\)
\(=\frac{\left(x-y\right)\left(1-x-y\right)}{x^2y^2+3}\)
\(=\frac{\left(x-y\right)\left[1-\left(x+4\right)\right]}{x^2y^2+3}\)
\(=\frac{\left(x-y\right).0}{x^2y^2+3}=0\)
Vậy : \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)
