Đáp án: $(x,y)\in\{(7,3),(\dfrac32,-\dfrac52)\}$
Giải thích các bước giải:
ĐKXĐ: $x-y>0$
Ta có : $x^2+y^2-4xy(\dfrac{2}{x-y}-1)=4(4+xy)$
$\to x^2+y^2-4xy\cdot \dfrac{2-(x-y)}{x-y}=16+4xy$
$\to x^2+y^2-4xy-16-4xy\cdot \dfrac{2-x+y}{x-y}=0$
$\to (x-y)^2-16-2xy-4xy\cdot \dfrac{2-x+y}{x-y}=0$
$\to (x-y)^2-4^2-2xy(1+2\cdot \dfrac{2-x+y}{x-y})=0$
$\to (x-y-4)(x-y+4)-2xy\cdot \dfrac{x-y+4-2x+2y}{x-y}=0$
$\to (x-y-4)(x-y+4)-2xy\cdot \dfrac{4-x+y}{x-y}=0$
$\to (x-y-4)(x-y+4)+2xy\cdot \dfrac{x-y-4}{x-y}=0$
$\to (x-y-4)(x-y+4+\dfrac{2xy}{x-y})=0$
$\to (x-y-4)((x-y)^2+4(x-y)+2xy)=0$
$\to (x-y-4)(x^2-2xy+y^2+4(x-y)+2xy)=0$
$\to (x-y-4)(x^2+y^2+4(x-y))=0$
$\to x-y-4=0\to x-y=4\to x=y+4$ vì $x-y>0\to x^2+y^2+4(x-y)>0$
Khi đó
$\sqrt{4}+3\sqrt{2y^2-y+1}=2y^2-(y+4)+3$
$\to 2+3\sqrt{2y^2-y+1}=2y^2-y-1$
$\to (2y^2-y+1)-3\sqrt{2y^2-y+1}-4=0$
$\to (\sqrt{2y^2-y+1}-4)(\sqrt{2y^2-y+1}+1)=0$
$\to \sqrt{2y^2-y+1}-4=0$ vì $\sqrt{2y^2-y+1}+1>0$
$\to \sqrt{2y^2-y+1}=4$
$\to 2y^2-y+1=16$
$\to 2y^2-y-15=0$
$\to (y-3)(2y+5)=0$
$\to y\in\{3,-\dfrac52\}$
$\to x\in\{7,\dfrac32\}$
