Question

\(\left\{{}\begin{matrix}xy+y^2=1+y\\x^2+2y^2+2xy=4+x\end{matrix}\right.\)

Answer 1

Lời giải:

Lấy PT(2) trừ 2PT(1) ta có:

$x^2=4+x-2(1+y)$
$\Leftrightarrow y=\frac{2+x-x^2}{2}$. Thay vào PT $(1)$:

$y(x+y)=y+1$

$\Leftrightarrow \frac{2+x-x^2}{2}.\frac{2+3x-x^2}{2}=\frac{4+x-x^2}{2}$

$\Leftrightarrow (2+x-x^2)(2+3x-x^2)=2(4+x-x^2)$

Đặt $2+x-x^2=a$ thì PT trở thành:

$a(a+2x)=6+a+x-x^2$

$\Leftrightarrow a^2+2ax+x^2-(a+x)-6=0$

$\Leftrightarrow (a+x)^2-(a+x)-6=0$

$\Leftrightarrow (a+x-3)(a+x+2)=0$

$\Leftrightarrow a+x-3=0$ hoặc $a+x+2=0$

Nếu $a+x-3=0$

$\Leftrightarrow 2x-x^2-1=0\Leftrightarrow x=1$

$\Rightarrow y=1$

Nếu $a+x+2=0$

$\Leftrightarrow 4+2x-x^2=0$

$\Leftrightarrow x=1\pm \sqrt{5}$

$\Rightarrow y=\frac{1}{2}(-3\mp \sqrt{5})$ (tương ứng)

Vậy..

Answer 2

https://i.imgur.com/9ayiA7v.png