a) Q = $\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)}$ - $\frac{\sqrt{x}+3}{\sqrt{x}-2}$ + $\frac{2\sqrt{x}+1}{\sqrt{x}-3}$
Q = $\frac{2\sqrt{x}-9-(\sqrt{x}+3)(\sqrt{x}-3)+(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}$
Q = $\frac{2\sqrt{x}-9-(x-9)+(2x-4\sqrt{x}+\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}$
Q = $\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}-3)}$
Q = $\frac{x-\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}-3)}$
Q = $\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}-3)}$
Q = $\frac{\sqrt{x}+1}{\sqrt{x}-3}$
b) Khi x = 1 ( thỏa mãn đkxđ ), ta có:
Q = $\frac{\sqrt{1}+1}{\sqrt{1}-3}$ = - 1
Vậy khi x = 1 thì Q = -1
c) Để Q = 2 thì $\frac{\sqrt{x}+1}{\sqrt{x}-3}$ = 2
<=> $\frac{\sqrt{x}+1}{\sqrt{x}-3}$ = $\frac{2(\sqrt{x}-3)}{\sqrt{x}-3}$
=> $\sqrt{x}$ + 1 = 2($\sqrt{x}$ - 3)
<=> $\sqrt{x}$ + 1 = 2$\sqrt{x}$ - 6
<=> $\sqrt{x}$ - 2$\sqrt{x}$ = -1- 6
<=> - $\sqrt{x}$ = - 7
<=> $\sqrt{x}$ = 7
<=> x = 49 (thỏa mãn ĐKXĐ)
Vậy để Q = 2 thì x = 49
