A) ĐKXĐ : \(x\ge0\:và\:x\ne4;9\)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\\ M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ M=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\Rightarrow M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\Rightarrow M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)B)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
M thuộc Z khi\(4⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3=-4;-2;-1;1;2;4\)
| \(\sqrt{x}-3\) | -4 | -2 | -1 | 1 | 2 | 4 |
| x | \(\varnothing\) | 1 | 4 | 16 | 25 | 49 |
vậy khi x={1;4;16;25;49} thì M thuộc Z
