giải hệ phương trình

2)

Đặt $\left\{ \begin{array}{l} \sqrt{x+2}=a\\ \sqrt{y+2}=b \end{array} \right.  a,b\geq 0 (*)$

hpt$\Leftrightarrow \left\{ \begin{array}{l} a^2-2-b=\frac{3}{2}\\ b^2-2+2(a^2-4)a=-\frac{7}{4} \end{array} \right.$

Tu $(1) :b=a^2-\frac{7}{2}(**)$ thay vao $(2):$ ta co

$(a^2-\frac{7}{2})^2+2a^3-8a-\frac{1}{4}=0$

$\Leftrightarrow a^4+2a^3-7a^2-8a+12=0$

$\Leftrightarrow (a+2)(a-2)(a-1)(a+3)=0$

$\Leftrightarrow a=1$ ,  $a=2$ ,$a=-2$(loai) hoac $a=-3$ (loai)ket hop voi $(**)$ta duoc $\left\{ \begin{array}{l} a=2\\ b=\frac{1}{2} \end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l} \sqrt{x+2}=2\\ \sqrt{y+2}=\frac{1}{2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=2\\ y=\frac{1}{2} \end{array} \right.$