a) Thay x=16 vào A ta có:
\(A=\dfrac{7}{\sqrt{16}+8}=\dfrac{7}{4+8}=\dfrac{7}{12}\)
b) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{x+3\sqrt{x}+3\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{\sqrt{x}+8}{\sqrt{x}-3}\left(dpcm\right)\)
