Đáp án:
a.$\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{x^2-2x+1}-\dfrac{1}{1-x^2})$$=-\dfrac{1+x}{x^2+1}$
b.$(\dfrac{x^2+xy}{x^3-x^2y+xy^2-y^3}+\dfrac{y}{x^2+y^2}):(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3})$$=x^2+y^2$
Giải thích các bước giải:
a.$\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{x^2-2x+1}-\dfrac{1}{1-x^2})$
$=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{(x-1)^2}+\dfrac{1}{x^2-1})$
$=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{(x-1)^2}+\dfrac{1}{(x-1)(x+1)})$
$=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{(x+1)^2}{(x-1)^2(x+1)^2}+\dfrac{(x-1)(x+1)}{(x-1)^2(x+1)^2})$
$=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{(x+1)^2+(x-1)(x+1)}{(x-1)^2(x+1)^2})$
$=\dfrac{1}{x-1}-\dfrac{x(x^2-1)}{x^2+1}.(\dfrac{(x+1).2x}{(x-1)^2(x+1)^2})$
$=\dfrac{1}{x-1}-\dfrac{x(x-1)(x+1)}{x^2+1}.(\dfrac{(x+1).2x}{(x-1)^2(x+1)^2})$
$=\dfrac{1}{x-1}-\dfrac{2x^2.(x+1)}{(x^2+1)(x-1)(x+1)}$
$=\dfrac{x^2+1}{(x-1)(x^2+1)}-\dfrac{2x^2}{(x^2+1)(x-1)}$
$=\dfrac{x^2+1-2x^2}{(x-1)(x^2+1)}$
$=\dfrac{1-x^2}{(x-1)(x^2+1)}$
$=\dfrac{(1-x)(1+x)}{(x-1)(x^2+1)}$
$=-\dfrac{1+x}{x^2+1}$
b.$(\dfrac{x^2+xy}{x^3-x^2y+xy^2-y^3}+\dfrac{y}{x^2+y^2}):(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3})$
$=(\dfrac{x(x+y)}{x^2(x-y)+y^2(x-y)}+\dfrac{y}{x^2+y^2}):(\dfrac{1}{x-y}-\dfrac{2xy}{x^2(x-y)+y^2(x-y)})$
$=(\dfrac{x(x+y)}{(x^2+y^2)(x-y)}+\dfrac{y(x-y)}{(x^2+y^2)(x-y)}):(\dfrac{1}{x-y}-\dfrac{2xy}{(x^2+y^2)(x-y)})$
$=\dfrac{x(x+y)+y(x-y)}{(x^2+y^2)(x-y)}:(\dfrac{x^2+y^2}{(x-y)(x^2+y^2)}-\dfrac{2xy}{(x^2+y^2)(x-y)})$
$=\dfrac{x^2+y^2}{(x^2+y^2)(x-y)}:\dfrac{x^2+y^2-2xy}{(x-y)(x^2+y^2)}$
$=\dfrac{1}{x-y}:\dfrac{(x-y)^2}{(x-y)(x^2+y^2)}$
$=\dfrac{1}{x-y}:\dfrac{x-y}{x^2+y^2}$
$=\dfrac{1}{x-y}.\dfrac{x^2+y^2}{x-y}$
$=x^2+y^2$
