1: Thay x=25 vào B, ta được:
\(B=\dfrac{25+7}{5}=\dfrac{32}{5}\)
2: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x-\sqrt{x}+6\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
1) \(B=\dfrac{x+7}{\sqrt{x}}=\dfrac{25+7}{\sqrt{25}}=\dfrac{32}{5}\)
2) \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-2x+\sqrt{x}+3}{x-9}=\dfrac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{x-9}=\dfrac{x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
