bài 1 tính phép tính sau: a)$\frac{1}{x-1}$ – $\frac{x^{3}- x}{x^2 +1}$ nhân ($\frac{1}{x^2 -2x+1}$ -$\frac{1}{1-x^2}$) b)$\frac{x^2 +xy}{x^3 +

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

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

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

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

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

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

$=\dfrac{1}{x-y}:\dfrac{x-y}{x^2+y^2}$

$=\dfrac{1}{x-y}.\dfrac{x^2+y^2}{x-y}$

$=x^2+y^2$