Đáp án`+`Giải thích các bước giải:

`a`, ($\frac{1}{x^2+x}$ - $\frac{2-x}{x+1}$ ):($\frac{1}{x}$ `x-2`) 

`=`($\frac{1}{x(x+1)}$ - $\frac{2x-x^2}{x(x+1)}$):( $\frac{1}{x}$ + $\frac{x^2}{x}$ - $\frac{2}{x}$) 

`=`$\frac{1-2x+x^2}{x(x+1)}$ : $\frac{1+x^2-2x}{x}$ 

`=`$\frac{1-2x+x^2}{x(x+1)}$ . $\frac{x}{1+x^2-2x}$ 

`=` $\frac{1}{x+1}$ 

`b`, ($\frac{3x}{1-3x}$ + $\frac{2x}{1+3x}$): $\frac{6x^2+10x}{(1-3x)^2}$ 

`=` ($\frac{3x+9x^2}{(1+3x)(1-3x)}$ + $\frac{2x-6x^2}{(1+3x)(1-3x)}$).$\frac{(1-3x)^2}{2x(3x+5)}$ 

`=` $\frac{3x+9x^2+2x-6x^2}{(1+3x)(1-3x)}$ .$\frac{(1-3x)^2}{2x(3x+5)}$ 

`=` $\frac{5x+3x^2}{(1+3x)(1-3x)}$ .$\frac{(1-3x)^2}{2x(3x+5)}$

`=`$\frac{x(3x+5)}{(1+3x)(1-3x)}$ .$\frac{(1-3x)^2}{2x(3x+5)}$

`=` $\frac{1-3x}{2(1+3x)}$ 

`c`, $\frac{x+1}{x+2}$ : ( $\frac{x+2}{x+3}$ . $\frac{x+1}{x+3}$) 

`=` $\frac{x+1}{x+2}$ : $\frac{(x+2)(x+1)}{(x+3)^2}$ 

`=` $\frac{x+1}{x+2}$ . $\frac{(x+3)^2}{(x+2)(x+1)}$ 

`=` $\frac{(x+3)^2}{(x+2)^2}$ 

`d`, $\frac{x^2-8xy+16y^2}{x^2-8xy+16y^2}$: $\frac{3x-12y}{3x+12y}$ 

`=` `1`:$\frac{3x-12y}{3x+12y}$ 

`=` `1`. $\frac{3x+12y}{3x-12y}$ 

`=` $\frac{3x+12y}{3x-12y}$ 

`=` $\frac{x+4y}{x-4y}$