1. $\left\{ \begin{array}{l} x^{4}-2x^{2}y+2y^{2}-1=0\\ x^{2} y^{2}-y^{3}+x^{2}-y^{2}-y+1 = 0\end{array} \right.$
2. $\left\{ \begin{array}{l} x-2y-\sqrt{xy}=0\\ \sqrt{x-1}+\sqrt{4y-1}=2 \end{array} \right.$
3. $\left\{ \begin{array}{l} x+y+\sqrt{x^{2}-y^{2}}=12\\ y\sqrt{x^{2}-y^{2}}=12 \end{array} \right.$
Bài 2:
Với $y=0=>x=0 (k là nghiệm) => y\neq 0$
Chia 2 vế Pt (1) cho y, ta được: $\frac{x}{y}-2-\frac{\sqrt{x}}{\sqrt{y}}=0$
Tìm được x theo y, rồi thế vào PT (2) để tìm x,y.
Bạn tự làm nốt nhé!
Bài 3.
Đặt: $a=y+\sqrt{x^{2}-y^{2}}=>a^{2}=x^{2}+2y\sqrt{x^{2}-y^{2}}=>\frac{a^{2}-x^{2}}{2}=y\sqrt{x^{2}-y^{2}}=12$
mà $a+x=12$
Bạn giải hệ tìm được a và x => y.
HPT (1) tương đương với:
$\begin{cases}(x^{2}-y)^{2}+(y^{2}-1)=0 \\ (x^{2}-y)(y^{2}-1)+2(x^{2}-y)-(y^{2}-1)=0 \end{cases}$
Bạn đổi biến rồi giải như bình thường. Bạn tự làm nốt nhé :D

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