Question

Answer 1

B1 đk x ≠ 9 , x ≥ 0

\(\text{a}=\dfrac{\sqrt{x}+1}{x-9}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-x\left(\sqrt{x}+3\right)}{\sqrt{x}\left(x-9\right)}=\dfrac{x+\sqrt{x}-x\sqrt{x}-3x}{\sqrt{x}\left(x-9\right)}=\dfrac{-2x+\sqrt{x}-x\sqrt{x}}{\sqrt{x}\left(x-9\right)}=\dfrac{\sqrt{x}\left(-2\sqrt{x}-x+1\right)}{\sqrt{x}\left(x-9\right)}=\dfrac{-x-2\sqrt{x}+1}{\left(x-9\right)}\)