GHPT:
1. $\left\{\begin{matrix}\sqrt{(1-y)(x^2+4x-8)} +(x+2)\sqrt{13-y}=12& \\ x^3+6x^2+4x=9+2\sqrt{y-3}& \end{matrix}\right.$
Điều kiện: $\left\{\begin{matrix} x> -2 & & \\ 3\leq y\leq 13 & & \end{matrix}\right.$
Giải quyết phương trình (1) ta có:
$\sqrt{\left ( 1-y \right )\left ( x^{2}+4x+8 \right )}= \sqrt{y-1}.\sqrt{8-4x-x^{2}}\leq \frac{y+7-4x-x^{2}}{2} \left ( \bigstar \right )$
$\sqrt{\left ( 13-y \right )\left ( x+2 \right )^{2}}\leq \frac{x^{2}+4x+17-y}{2}\left ( \bigstar \bigstar \right )$
Từ $\left ( \bigstar \right );\left ( \bigstar \bigstar \right )\rightarrow VT_{\left ( 1 \right )}\leq 12$
Dấu ''='' xảy ra $\Leftrightarrow y=-x^{2}-4x+9$