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a) ĐKXĐ:  \(x>0;x\ne9\)

\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\frac{1}{\sqrt{x}+3}\)

b)  \(A=\frac{1}{5}\) \(\Rightarrow\)\(\frac{1}{\sqrt{x}+3}=\frac{1}{5}\)

\(\Rightarrow\)\(\sqrt{x}+3=5\)

\(\Leftrightarrow\)\(\sqrt{x}=2\)

\(\Leftrightarrow\)\(x=4\)(t/m ĐKXĐ)

Vậy...

ĐK: \(x\ge\frac{2}{3}\)

\(pt\Leftrightarrow5\sqrt{4x+1}-5\sqrt{3x-2}=4x+1-\left(3x-2\right)\)

Đặt \(a=\sqrt{4x+1};\text{ }b=\sqrt{3x-2}\text{ }\left(a;\text{ }b\ge0\right)\)

Pt trở thành: \(5a-5b=a^2-b^2\Leftrightarrow\left(a-b\right)\left(a+b\right)-5\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-5\right)=0\)\(\Leftrightarrow a=b\text{ hoặc }a+b=5\)

\(+\text{Nếu }a=b\text{ thì }\sqrt{4x+1}=\sqrt{3x-2}\Leftrightarrow4x+1=3x-2\Leftrightarrow x=-3\text{ }\left(\text{loại}\right)\)

\(+\text{Nếu }a+b=5\text{ thì }\sqrt{4x+1}+\sqrt{3x-2}=5\)

\(\Leftrightarrow4x+1+3x-2+2\sqrt{\left(4x+1\right)\left(3x-2\right)}=25\)

\(\Leftrightarrow2\sqrt{12x^2-5x-2}=26-7x\)

\(\Leftrightarrow4\left(12x^2-5x-2\right)=\left(26-7x\right)^2\text{ và }26-7x\ge0\)

\(\Leftrightarrow x^2-344x+684=0\text{ và }x\le\frac{26}{7}\)

\(\Leftrightarrow\left(x-342\right)\left(x-2\right)=0\text{ và }x\le\frac{26}{7}\)

\(\Leftrightarrow x=342\text{ hoặc }x=2\text{ và }x\le\frac{26}{7}\)

\(\Leftrightarrow x=2\)

\(\text{Kết luận: }x=2.\)