Đkxđ : \(x\ne4\); \(x\ge0\)

a.\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\left(\sqrt{x}\right)^2+3\sqrt{x}-4\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

b. Ta có : \(A=\) \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(\Leftrightarrow A\left(\sqrt{x}-2\right)=\sqrt{x}-4\)

\(\Leftrightarrow A\sqrt{x}-2A=\sqrt{x}-4\)

\(\Leftrightarrow A\sqrt{x}-\sqrt{x}=2A-4\)

\(\Leftrightarrow\sqrt{x}\left(A-1\right)=2\left(A-2\right)\)

\(\Leftrightarrow\sqrt{x}=\dfrac{2\left(A-2\right)}{A-1}\)

\(\Leftrightarrow x=\left(\dfrac{2\left(A-2\right)}{A-1}\right)^2\)

Vậy .....

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