cho x+y =1 va x.y =0 cmr: $\frac{x}{y^3-1}$ – $\frac{y}{x^3-1}$ + $\frac{2.(x-y)}{x^2y^2 +3}$ =0

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1. Giải thích các bước giải:

   Ta có : 

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

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

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

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

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

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

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

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

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

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

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

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

   $=0$

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