Giải hệ phương trình: $\ \left\{ \begin{array}{l} {x^3} = \sqrt {4 - {x^2}} + 2\sqrt y \\ 3{x^4} + 4y = 2x\sqrt y \left( {{x^2} + 3} \right) \end{array} \right.,\,\left( {x,y \in R} \right).$

Đề bài: (Đề thi thử ĐH số 1 - Báo THTT số 435/T9.2013)
Giải hệ phương trình: $\ \left\{ \begin{array}{l}
{x^3} = \sqrt {4 - {x^2}}  + 2\sqrt y \\
3{x^4} + 4y = 2x\sqrt y \left( {{x^2} + 3} \right)
\end{array} \right.,\,\left( {x,y \in R} \right).$
Giải:
HPT $\ \Leftrightarrow \left\{ \begin{array}{l}
2\sqrt y  = {x^3} - \sqrt {4 - {x^2}} \\
3{x^4} + {\left( {{x^3} - \sqrt {4 - {x^2}} } \right)^2} = \left( {{x^2} + 3} \right)\left( {{x^4} - x\sqrt {4 - {x^2}} } \right)\left( * \right)
\end{array} \right.$

$\ \left( * \right) \Leftrightarrow 3{x^4} + {x^6} - 2{x^3}\sqrt {4 - {x^2}}  + 4 - {x^2} = {x^6} - {x^3}\sqrt {4 - {x^2}}  + 3{x^4} - 3x\sqrt {4 - {x^2}} .$
$\  \Leftrightarrow {x^3}\sqrt {4 - {x^2}}  - 3x\sqrt {4 - {x^2}}  = {\left( {\sqrt {4 - {x^2}} } \right)^2}.$
$\  \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 - {x^2} = 0\\
x > 0
\end{array} \right. \Rightarrow \left( {x;y} \right) = \left( {2;16} \right)\\
\left\{ \begin{array}{l}
{x^3} - 3x = \sqrt {4 - {x^2}} \\
x > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2cost\left( {t \in \left( { - \frac{\pi }{2};\frac{\pi }{2}} \right)} \right)\\
2\left( {4co{s^3}t - 3\cos t} \right) = 2\sin t
\end{array} \right.\left( {**} \right)
\end{array} \right.$
$\ \left( {**} \right) \Leftrightarrow \left\{ \begin{array}{l}
x = 2cost\left( {t \in \left( {0;\frac{\pi }{2}} \right)} \right)\\
cos3t = cos\left( {\frac{\pi }{2} - t} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2cost\left( {t \in \left( { - \frac{\pi }{2};\frac{\pi }{2}} \right)} \right)\\
t = \frac{\pi }{8};\frac{\pi }{4};\frac{{3\pi }}{8}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left( {x;y} \right) = \left( {2cos\frac{\pi }{8};9co{s^2}\frac{\pi }{8}} \right)\\
\left( {x;y} \right) = \left( {2cos\frac{\pi }{4};9co{s^2}\frac{\pi }{4}} \right)\\
\left( {x;y} \right) = \left( {2cos\frac{{3\pi }}{8};9co{s^2}\frac{{3\pi }}{8}} \right)
\end{array} \right.$
  Vậy $\ S = \left\{ {\left( {2;16} \right);\left( {2cos\frac{\pi }{8};9co{s^2}\frac{\pi }{8}} \right);\left( {2cos\frac{\pi }{4};9co{s^2}\frac{\pi }{4}} \right);\left( {2cos\frac{{3\pi }}{8};9co{s^2}\frac{{3\pi }}{8}} \right)} \right\}.$