ĐKXĐ: \(x\ge0;x\ne\left\{4;9\right\}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)
\(\Rightarrow A=\frac{2-\sqrt{3}+1}{2-\sqrt{3}-3}=3-2\sqrt{3}\)
\(A=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để A nguyên \(\Rightarrow\sqrt{x}-3=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1\left(ktm\right);1;2;4;5;7\right\}\)
\(\Rightarrow x=\left\{1;4\left(ktm\right);16;25;49\right\}\)
