1, \begin{cases}\frac{17-x^{2}}{y}=\sqrt{x}(3+\sqrt{x})+2\sqrt{63-14x-18y} \\ x(x^{2}+2x+9)+12y=34+2(13-13y)\sqrt{17-6y} \end{cases}
ĐK : ....
pt (2) $<=> x^{3}+2x^{2}+9x=(17-6y)\sqrt{17-6y}+2(17-6y)+9\sqrt{17-6y}$
Xét hàm số : $f(t)=t^{3}+2t^{2}+9t=>f(t')=3t^{2}+4t+9=(t+2)^{2}+2t^{2}+5>0$              $(\forall  t \in R)$
$=> $ Hàm số liên tục và đồng biến trên $R$ :
  Mà : 
    $f(x)=f(\sqrt{17-6y})=> \begin{cases}x^{2}=17-6y \\ x \geq 0\end{cases}$
$=>$ Pt (1) : 
 $6y/y=3\sqrt{x}+x+2\sqrt{63-14x+3(x^{2}-17)}$
$<=>4+(2-x)-3\sqrt{x}=2\sqrt{3(2-x)^{2}-2x}$
Tới chỗ này đặt :  $\begin{cases}2-x=u \\ \sqrt{x}=v \end{cases}$ Rồi giải pt là ok 
chắc lúc đó mk bấm lộn –  ☼SunShine❤️ 21-05-16 05:10 PM
chỗ mt^2 thay m bằng 3 đi bạn!!! Đạo hàm mà!!! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 21-05-16 05:09 PM
bài nào ? –  ☼SunShine❤️ 03-04-16 12:49 PM
oh....mà gửi cho tui cái link bài đó! –  Confusion 03-04-16 12:49 PM
cái này học thêm thôi -_- thầy tuôi gt –  ☼SunShine❤️ 03-04-16 12:47 PM
đã hk đạo hàm r ak? –  Confusion 03-04-16 12:43 PM

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