$\begin{cases}9x^3+x-(y-\frac{5}{3})\sqrt{3y-6}=0 \\ x^2+x+2=\sqrt{y+2} \end{cases}$
có sai đề j ko bạn sao kết quả lẻ thế –  ≧◔◡◔≦ ۩๖ۣۜNguyễn's Đức♫10x۩ 15-07-16 03:02 PM
ĐKXĐ: $y\geq 2$
Xét PT $(1)$ ta có: 
$Pt\Leftrightarrow 27x^3+3x-(3y-5)\sqrt{3y-6}=0\Leftrightarrow 27x^3+3x=(3y-5)\sqrt{3y-6}$
$\Leftrightarrow (3x)^3+3x=(3y-6)\sqrt{3y-6}+\sqrt{3y-6}$
Xét hàm số $f(t)=t^3+t\rightarrow f'(t)=3t^2+1>0$ $\Rightarrow $ Hàm số đồng biến trên D
Do đó : $Pt (1) \Leftrightarrow 3x=\sqrt{3y-6}\Rightarrow \begin{cases}x\geq 0\\ 9x^2=3y-6 \end{cases}$
$y=3x^2+2$
Thế vào $(2)$ ta có :$x^2+x+2=\sqrt{3x^2+4}$
Đến đây bạn bình phương tự giải hoặc nhân liên hợp (TÙY Nhé)
VOte mạnh giúp –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 15-07-16 05:11 PM

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