`Q=(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2})(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1})`
`a)`ĐKXĐ: `a>=0`
`->\sqrt{a}-2\ne0<=>a\ne4`
`Q=(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2})(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1})(a>=0,a\ne4)`
`=(3+\frac{\sqrt{a}(\sqrt{a}-2)}{\sqrt{a}-2})(3-\frac{\sqrt{a}(3\sqrt{a}+1)}{3\sqrt{a}+1})`
`=(3+\sqrt{a})(3-\sqrt{a})`
`=3^2-\sqrt{a}^2`
`=9-a`
`b)a=\sqrt{6-4\sqrt{2}}=\sqrt{(2-\sqrt{2})^2}=2-\sqrt{2}`
`->Q=9-a=9-(2-\sqrt{2})=9-2+\sqrt{2}=7+\sqrt{2}`
Vậy `a=\sqrt{6-4\sqrt{2}}` thì `Q=7+\sqrt{2}`
