Đáp án + Giải thích các bước giải:

`a)`  $A=\dfrac{4xy}{y^2-x^2}:\bigg(\dfrac{1}{y^2-x^2}+\dfrac{1}{y^2-2xy+x^2}\bigg)$

ĐKXĐ: $\begin{cases} y^2-x^2\ne0\\y^2-2xy+x^2\ne0\\\dfrac{1}{y^2-x^2}+\dfrac{1}{y^2-2xy+x^2}\ne0 \end{cases}$

$⇔\begin{cases} (y-x)(y+x)\ne0\\(y-x)^2\ne0\\\dfrac{1}{(y-x)(y+x)}+\dfrac{1}{(y-x)^2}\ne0 \end{cases}$

$⇔\begin{cases} x\ne\pm y\\\dfrac{y-x+y+x}{(y-x)^2(y+x)}\ne0 \end{cases}$

$⇔\begin{cases} x\ne\pm y\\\dfrac{2y}{(y-x)^2(y+x)}\ne0 \end{cases}$

$⇔\begin{cases} x\ne\pm y\\y\ne0 \end{cases}$

`b)` $A=\dfrac{4xy}{y^2-x^2}:\bigg(\dfrac{1}{y^2-x^2}+\dfrac{1}{y^2-2xy+x^2}\bigg)$

$=\dfrac{4xy}{(y-x)(y+x)}:\bigg[\dfrac{1}{(y-x)(y+x)}+\dfrac{1}{(y-x)^2}\bigg]$

$=\dfrac{4xy}{(y-x)(y+x)}:\dfrac{y-x+y+x}{(y-x)^2(y+x)}$

$=\dfrac{4xy}{(y-x)(y+x)}.\dfrac{(y-x)^2(y+x)}{2y}$

$=2x(y-x)$

$=2xy-2x^2$

`#Lighth`

`a)` Đkxđ: `x ne ±y; y ne 0`

`b)`

`A = (4xy)/(y^2 - x^2) : ( 1/(y^2 - x^2) + 1/(y^2 + 2xy + x^2) )`

`A = (4xy)/(y^2 - x^2) : (1/((y - x)(y + x)) + 1/((y + x)^2) )`

`A = (4xy)/(y^2 - x^2) : ( (y + x)/((y - x)(y + x)^2 + (y - x)/((y - x)(y + x)^2) )`

`A = (4xy)/(y^2 - x^2) : (y + x + y - x)/((y - x)(y + x)^2)`

`A = (4xy)/(y^2 - x^2) . ((y - x)(y + x)^2)/(2y)`

`A = (4xy(y + x)/(2y)`

`A = 2x(x + y)`