Giải thích các bước giải:

Câu 18:

Ta có:

$x-2y^2-y+1=0$

$\to 2y^2=x-y+1$

$\to x^3-xy^2-2y^2(x+y)=0$

$\to x^3-xy^2-2xy^2-2y^3=0$

$\to x^3-3xy^2-2y^3=0$

$\to (x+y)^2(x-2y)=0$

$\to x=-y$ hoặc $x=2y$

Giải $x=-y$

$\to -y-2y^2-y+1=0$

$\to -2y^2-2y+1=0$

$\to 2y^2+2y-1=0$

$\to y=\dfrac{-1\pm\sqrt3}2$

$\to x=-\dfrac{-1\pm\sqrt3}2$

Giải $x=2y$

$\to 2y-2y^2-y+1=0$

$\to -2y^2+y+1=0$

$\to -(2y-1)(y-1)=0$

$\to y\in\{\dfrac12, 1\}$

$\to x\in\{1, 2\}$

Câu 19:

Ta có:

$x^2+1=xy$

$\to x^2-xy+1=0$

$\to 4x^2-4xy+4=0$

$\to 4x^2-4xy+y^2+4-y^2=0$

$\to (2x-y)^2+4-y^2=0$

$\to (2x-y)^2+4-y^2+\sqrt{x-2y-3}+2y^2+4y=0$

$\to (2x-y)^2+\sqrt{x-2y-3}+y^2+4y+4=0$

$\to (2x-y)^2+\sqrt{x-2y-3}+(y+2)^2=0$

Mà $(2x-y)^2+\sqrt{x-2y-3}+(y+2)^2\ge 0$

$\to 2x-y=x-2y-3=y+2=0$

$\to x=-1, y=-2$