Giải thích các bước giải:
a.Ta có :
$\begin{cases}x^2+y^2=2x^2y^2\\(x+y)(1+xy)=4x^2y^2\end{cases}$
$\to \begin{cases}x^2+2xy+y^2=2x^2y^2+2xy\\(x+y)(1+xy)=4x^2y^2\end{cases}$
$\to \begin{cases}(x+y)^2=2xy(xy+1)\\(x+y)(1+xy)=4x^2y^2\end{cases}$
$\to (x+y)^2-4x^2y^2=2xy(xy+1)-(x+y)(1+xy)$
$\to (x+y-2xy)(x+y+2xy)=(2xy-(x+y))(xy+1)$
$\to (x+y-2xy)(x+y+2xy)+((x+y)-2xy)(xy+1)=0$
$\to (x+y-2xy)(x+y+2xy+xy+1)=0$
$\to (x+y-2xy)(x+y+3xy+1)=0$
$+)x+y-2xy=0\to x+y=2xy\to 2xy(1+xy)=4x^2y^2$
$\to xy=0\to x=y=0$
Hoặc $1+xy=2xy\to xy=1\to x+y=2$
Mà $(x+y)^2=4xy\to x=y=1$
$+) x+y+3xy+1=0\to x+y=-(3xy+1)$
$\to -(3xy+1)(xy+1)=4x^2y^2$
$\to 7x^2y^2+4xy+1=0\to$ vô nghiệm
b.Ta có : $2x+y=2\to y=2-2x$
$\to x^2+2(2-2x)+\sqrt{x^2+2(2-2x)}=1$
$\to x^2-4x+4+\sqrt{x^2-4x+4}=1$
$\to x^2-4x+4+\sqrt{(x-2)^2}=1$
$\to x^2-4x+4+|x-2|=1$
$+)x\ge 2\to x^2-4x+4+x-2=1\to x=\dfrac{3+\sqrt{5}}{2},\:x=\dfrac{3-\sqrt{5}}{2}$
$\to x=\dfrac{3+\sqrt{5}}{2}$
$+)x<2\to x^2-4x+4-x+2=1\to \:x=\dfrac{5-\sqrt{5}}{2}$
