Giải thích các bước giải:
Ta có :
$x^2-y^2-4x=2y-3$
$\to x^2-4x+4=y^2+2y+1$
$\to (x-2)^2=(y+1)^2$
$\to x-2=y+1\to y=x-3\to y-x+1=-2<0\to Loại$
$\to x-2=-(y+1)\to y=-x+1$
$\to \sqrt{y-x+1}+2=\sqrt{3x+1}$
$\to \sqrt{-x+1-x+1}+2=\sqrt{3x+1}$
$\to \sqrt{-2x+2}+2=\sqrt{3x+1}$
$\to \left(\sqrt{-2x+2}+2\right)^2=\left(\sqrt{3x+1}\right)^2$
$\to -2x+4\sqrt{-2x+2}+6=3x+1$
$\to 4\sqrt{-2x+2}=5x-5$
$\to \left(4\sqrt{-2x+2}\right)^2=\left(5x-5\right)^2$
$\to \left(4\sqrt{-2x+2}\right)^2=\left(5x-5\right)^2$
$\to -32x+32=25x^2-50x+25$
$\to 25x^2-18x-7=0$
$\to x=1\to y=0,x=-\dfrac{7}{25}\to y=\dfrac{32}{25}$
