1. \begin{cases}x(x+y+1)-3=0 \\ (x+y)^{2}-\frac{5}{x^{2}}+1=0 \end{cases}
2. \begin{cases}xy+x+y=x^{2}-2y^{2} \\ x\sqrt{2y}-y\sqrt{x-1}= 2(x-y)\end{cases}
pt1: x(x+y+1)-3=0 <=> x+y=$\frac{3}{x}-1$
thê vào pt 2 ta có: $(\frac{3}{x}-1)^{2}$-$\frac{5}{x^{2}}$+1=0 khai triển pt này ra e tìm đk x nha
Câu 2:
PT(1):$xy+y^2+x+y=x^2-y^2$
$\Leftrightarrow y(x+y)+(x+y)=(x-y)(x+y)$
$\Leftrightarrow (x+y)(y+1)=(x-y)(x+y)$
Từ đây $2y=x-1$ hoặc $x=-y$
Câu 1:$ĐK:x\neq0$
\begin{cases}x(x+y)+x-3=0 \\ (x+y)^2-\frac{5}{x^2}+1=0 \end{cases}
$\Leftrightarrow \begin{cases}x+y= \frac{3-x}{x}\\ (\frac{3}{x}-1)^2-\frac{5}{x^2}+1=0(1) \end{cases}$
$(1)\Leftrightarrow \frac{9}{x^2}-\frac{6}{x}-\frac{5}{x^2}+2=0$
Từ đó ta tìm đc x=1 hoặc $x=\pm \sqrt{2}$

hệ đã cho tương đương với 
$ \begin{cases}x(x+y)+x-3=0 \\ x^{2}(x+y)^{2}+x^{2}-5=0 \end{cases}$
đặt $ x(x+y)=a$ ra hệ dễ ợt

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