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

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 $x+y=1$

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

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

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

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

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

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

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

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

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

$=0$