Câu a : ĐKXĐ : \(x\ge0\) và \(x\ne9\)
\(B=\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
Câu b : \(A=\left(\sqrt{8}-\sqrt{12}\right)\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\sqrt{16}+\sqrt{24}-\sqrt{24}-\sqrt{36}\)
\(=\sqrt{16}-\sqrt{36}\)
\(=4-6=-2\)
Câu c : Để : \(A=B\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-2\)
\(\Leftrightarrow2\sqrt{x}=-2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow2\sqrt{x}=-2\left(x-9\right)\)
\(\Leftrightarrow2x+2\sqrt{x}-18=0\)
Tới khúc này chịu