Câu $4.\left\{ \begin{array}{l} x\sqrt{12-y}+\sqrt{y(12-x^2)}=12\\ bla.....bla..... \end{array} \right.$
Xài $Cauchy-Schwarz:$
$VT_{(1)}\leq \sqrt{(x^2+12-x^2)(12-y+y)}=12$
Đẳng thức khi:
$\frac{x}{\sqrt{12-y}}=\frac{\sqrt{12-x^2}}{\sqrt{y}}\Leftrightarrow \left\{ \begin{array}{l} 0\leq x\leq 2\sqrt{3}\\ x\sqrt{y}=\sqrt{12-y}.\sqrt{12-x^2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x \in.......\\ x^2y=(12-y)(12-x^2) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x \in........\\ x^2y=144-12x^2-12y+x^2y \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\in.........\\ y=12-x^2 \end{array} \right.$
Thế zô $(2)......$ ~~đang tính toán~~
$\Leftrightarrow (x-3)[x^2+3x+1+\frac{2(x+3)}{1+\sqrt{10-x^2}}]=0$
$\rightarrow (x;y)=(3;3)$