`~rai~`
\(B=\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\\a)ĐKXĐ:\begin{cases}x\ge 0\\2-\sqrt{x}\ne 0\\x+\sqrt{x}-6\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ge 0\\x\ne 4\end{cases}\\\text{Với }x\ge 0;x\ne 4,\text{ta có:}\\B=\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\\\quad=\dfrac{(3-\sqrt{x})(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}+3)}+\dfrac{(\sqrt{x}-2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+3)}+\dfrac{x-9}{(\sqrt{x}-2)(\sqrt{x}+3)}\\\quad=\dfrac{9-x+x-4\sqrt{x}+4+x-9}{(\sqrt{x}-2)(\sqrt{x}+3)}\\\quad=\dfrac{x-4\sqrt{x}+4}{(\sqrt{x}-2)(\sqrt{x}+3)}\\\quad=\dfrac{(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\\\quad=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}.\\\text{Vậy với }x\ge 0;x\ne 4\text{ thì B}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}.\\b)\text{Ta có:}x=9-4\sqrt{5}\\\Leftrightarrow \sqrt{x}=\sqrt{9-4\sqrt{5}}\\\Leftrightarrow \sqrt{x}=\sqrt{5-4\sqrt{5}+4}\\\Leftrightarrow \sqrt{x}=\sqrt{(\sqrt{5}-2)^2}\\\Leftrightarrow \sqrt{x}=|\sqrt{5}-2|\\\Leftrightarrow \sqrt{x}=\sqrt{5}-2.\quad\text{(vì }\sqrt{5}>2)\\\text{Thay }\sqrt{x}=\sqrt{5}-2\text{ vào biểu thức B được:}\\B=\dfrac{\sqrt{5}-2-2}{\sqrt{5}-2+3}\\\quad=\dfrac{\sqrt{5}-4}{\sqrt{5}+1}\\\quad=\dfrac{(\sqrt{5}-4)(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\\\quad=\dfrac{5-\sqrt{5}-4\sqrt{5}+4}{5-1}\\\quad=\dfrac{9-5\sqrt{5}}{4}.\\\text{Vậy với }x=9-4\sqrt{5}\text{ thì giá trị biểu thức B là }\dfrac{9-5\sqrt{5}}{4}.\\c)B=\dfrac{7}{12}\\\Leftrightarrow \dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{7}{12}\\\Leftrightarrow 12(\sqrt{x}-2)=7(\sqrt{x}+3)\\\Leftrightarrow 12\sqrt{x}-24=7\sqrt{x}+21\\\Leftrightarrow 12\sqrt{x}-7\sqrt{x}=21+24\\\Leftrightarrow 5\sqrt{x}=45\\\Leftrightarrow \sqrt{x}=9\\\Leftrightarrow x=81.\text{(thỏa mãn ĐKXĐ)}\\\text{Vậy với }x=81\text{thì B}=\dfrac{7}{12}.\)
