Question

Giải hệ \(\left\{{}\begin{matrix}x-y-z=2\left(\sqrt{yz}+\sqrt{y}+\sqrt{z}-\sqrt{x}\right)\\3\sqrt{yz}=x-\sqrt{3z}+1\end{matrix}\right.\)

Answer 1

Giúp em với ạ Akai Haruma

Answer 2

Lời giải:

PT $(1)$ tương đương với:

$x+2\sqrt{x}+1=y+z+2\sqrt{yz}+2\sqrt{y}+2\sqrt{z}+1$

$\Leftrightarrow (\sqrt{x}+1)^2=(\sqrt{y}+\sqrt{z}+1)^2$

\(\left[\begin{matrix} \sqrt{x}=\sqrt{y}+\sqrt{z}\\ \sqrt{x}=-(\sqrt{y}+\sqrt{z})\end{matrix}\right.\)

Nếu $\sqrt{x}=-(\sqrt{y}+\sqrt{z})$

$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}=0\Rightarrow x=y=z=0$ (không thỏa mãn PT $(2)$)

Nếu $\sqrt{x}=\sqrt{y}+\sqrt{z}$

$\Rightarrow 3\sqrt{yz}=(\sqrt{y}+\sqrt{z})^2-\sqrt{3z}+1$

$\Leftrightarrow \sqrt{yz}=y+z-\sqrt{3z}+1$

$\Leftrightarrow 4y+4z-4\sqrt{yz}-4\sqrt{3z}+4=0$

$\Leftrightarrow (2\sqrt{y}-\sqrt{z})^2+(\sqrt{3z}-2)^2=0$

$\Rightarrow (2\sqrt{y}-\sqrt{z})^2=(\sqrt{3z}-2)^2=0$

$\Rightarrow z=\frac{4}{3}; y=\frac{1}{3}; x=3$