a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\\sqrt{x}+3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
- Ta có : \(B=\frac{1}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}\)
=> \(B=\frac{\sqrt{x}+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\)
=> \(B=\frac{\sqrt{x}+3+\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}}{x-9}\)
b, Ta có : \(A=\left(\sqrt{8}-\sqrt{12}\right)\left(\sqrt{2}+\sqrt{3}\right)\)
=> \(A=\sqrt{16}-\sqrt{24}+\sqrt{24}-\sqrt{36}\)
=> \(A=\sqrt{16}-\sqrt{36}=4-6=-2\)
c, - Để A = B khi : \(\frac{2\sqrt{x}}{x-9}=-2\)
=> \(2\sqrt{x}=18-2x\)
=> \(2x+2\sqrt{x}-18=0\)
=> \(\left(\sqrt{2x}\right)^2+2\sqrt{2x}.\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{37}{2}\)
=> \(\left(\sqrt{2x}+\frac{1}{\sqrt{2}}\right)^2=\frac{37}{2}\)
=> \(\left[{}\begin{matrix}x=\left(\frac{\sqrt{\frac{37}{2}}-\frac{1}{\sqrt{2}}}{\sqrt{2}}\right)^2=\frac{19-\sqrt{37}}{2}\\x=\left(\frac{-\sqrt{\frac{37}{2}}-\frac{1}{\sqrt{2}}}{\sqrt{2}}\right)^2=\frac{19+\sqrt{37}}{2}\end{matrix}\right.\)
Vậy để A = B thì \(\left[{}\begin{matrix}x=\frac{19-\sqrt{37}}{2}\\x=\frac{19+\sqrt{37}}{2}\end{matrix}\right.\)
