√9-x² - x + 3 = 0

`\sqrt{9-x^2}-x+3=0` `(-3<=x<=3)`

`<=>\sqrt{(3-x)(3+x)}+(3-x)=0`

`<=>\sqrt{3-x}.\sqrt{3+x}+(\sqrt{3-x})^2=0`

`<=>\sqrt{3-x}.(\sqrt{3+x}+\sqrt{3-x})=0`

`<=>`\(\left[ \begin{array}{l}\sqrt{3-x}=0\\\sqrt{3+x}+\sqrt{3-x}=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\sqrt{3-x}=0\\\sqrt{3-x}=-\sqrt{3+x} \text{(vô lý)}\end{array} \right.\) 

`<=>\sqrt{3-x}=0`

`<=>3-x=0`

`<=>-x=-3`;

`<=>x=3` `(tm)`

Vậy `S={3}`

Giải thích các bước giải+Đáp án:

`\sqrt{9-x^2}-x+3=0`    `(x>=3)` hoặc `(-3<=x<=3)`

`<=>\sqrt{9-x^2}-(x-3)=0`

`<=>\sqrt{(3-x)(3+x)}-(\sqrt{x-3})^2=0`

`<=>\sqrt{-(x-3)(3+x)}-\sqrt{x-3}.\sqrt{x-3}=0`

`<=>\sqrt{x-3}(\sqrt{-(3+x)}-\sqrt{x-3})=0`

`<=>`\(\left[ \begin{array}{l}\sqrt{x-3}=0\\\sqrt{-(3+x)}-\sqrt{x-3}=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x-3=0\\\sqrt{-(3+x)}=\sqrt{x-3}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=3\\-3-x=x-3\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=3\\x=0(KTM)\end{array} \right.\) 

 Vậy: `S={3}`