Giải thích các bước giải:
ĐKXĐ : $-1\le \:x\le \dfrac{1}{2}$
Ta có : $x(2x-2y+1)=y$
$\to 2x^2-2xy+x=y$
$\to 2x^2+x=y+2xy$
$\to x(2x+1)=y(1+2x)$
$\to (x-y)(2x+1)=0$
$+)2x+1=0\to x=-\dfrac{1}{2}$
$\to y+2\sqrt{1-x-2x^2}=2(1+y)^2$
$\to y+2\sqrt{1-\left(-\dfrac{1}{2}\right)-2\left(-\dfrac{1}{2}\right)^2}=2\left(1+y\right)^2$
$\to y+2=2+4y+2y^2$
$\to 2y^2+3y=0$
$\to y\in\{0,-\dfrac{3}{2}\}$
$+)x=y$
$\to y+2\sqrt{1-x-2x^2}=2(1+y)^2$
$\to x+2\sqrt{1-x-2x^2}=2(1+x)^2$
$\to 2\sqrt{1-x-2x^2}=2x^2+3x+2$
$\to \left(2\sqrt{1-x-2x^2}\right)^2=\left(2x^2+3x+2\right)^2$
$\to 4-4x-8x^2=4x^4+12x^3+17x^2+12x+4$
$\to x\left(4x^3+12x^2+25x+16\right)=0$
$\to x=0\to y=0$
