Điều kiện: x \(\ge\)0; x \(\ne\) 4;x \(\ne\) 9
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}-2\right)}\)
\(A=\frac{2\sqrt{x}-9-\left(x-9\right)+\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}\)
\(A=\frac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để A nguyên thì \(\frac{4}{\sqrt{x}-3}\) nguyên <=> \(\sqrt{x}-3\) \(\in\)Ư(4) = {4;-4;2;-2;1;-1}
| \(\sqrt{x}-3\) | 4 | -4 | 2 | -2 | 1 | -1 |
| \(\sqrt{x}\) | 7 | -1 | 5 | 1 | 4 | 2 |
| x | 49 | loại | 25 | 1 | 16 | 4 |
Đối chiếu điều kiện => x \(\in\) {49;25;1;16}
