a
\(A=\dfrac{\sqrt{9}-2}{\sqrt{9}+2}=\dfrac{3-2}{3+2}=\dfrac{1}{5}\)
b
\(B=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{7\sqrt{x}+1}{x-2\sqrt{x}+3\sqrt{x}-6}\\ =\dfrac{2x+6\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{7\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)+3\left(\sqrt{x}-2\right)}\\ =\dfrac{2x+5\sqrt{x}-3-x+2\sqrt{x}-7\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
c cắt quá không nhìn được bạn: )
