GHPT

Question

$\left\{ \begin{array}{l} 2\sqrt{x+3y+2}-3\sqrt{y}= \sqrt{x+2}\\ \sqrt{y-1}-\sqrt{4-x}+8-x^2=0\end{array} \right.$

Nero · Answer 1 · 12/5/2014 10:16:43 PM

Đk:

$\begin{cases}-2\leq x\leq 4\\ y\geq 1 \\ x+3y+2\geq 0\end{cases}$

Pt

$(1): 2\sqrt{x+3y+2}=\sqrt{x+2}+3\sqrt{y}$

Bình phương 2 vế ta được:

$4(x+3y+2)=x+3y+2+6y+6\sqrt{y(x+2)}$

$\Leftrightarrow x+2-2\sqrt{y(x+2})+y=0$

$\Leftrightarrow (\sqrt{x+2}-\sqrt{y})^2=0$

$\Leftrightarrow\sqrt{x+2}=\sqrt{y}$

$\Leftrightarrow x+2=y$

Thế vào

$(2)\Leftrightarrow \sqrt{x+1}-\sqrt{4-x}+8-x^2=0$

$\Leftrightarrow \sqrt{x+1}-2+1-\sqrt{4-x}+9-x^2=0$

$\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{4-x}+1}-(x-3)(x+3)=0$

$\Leftrightarrow x=3\vee \frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{4-x}+1}-x-3=0$

$(\bigstar)$

$(\bigstar)<0\Rightarrow (\bigstar)$ vô nghiệm

Từ đó

$x=3\Rightarrow y=5$

1 Đáp án