Đáp án:
a, A=$\frac{\sqrt{x}(\sqrt{x}+2)}{x\sqrt{x}+1}$
b, A=$\frac{8}{9}$
Giải thích các bước giải:
A=$(\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)} -\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}))+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)} ).\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}$
=$(\frac{2\sqrt{x}-9-x+9+2x-\sqrt{x}-2}{(\sqrt{x}-2)}). (\frac{\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)})$
=$\frac{x-2}{\sqrt{x}-2} .\frac{\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}$
=$\frac{\sqrt{x}(\sqrt{x}+2)}{x\sqrt{x}+1}$
b, $x=\sqrt{3+2.\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-6\sqrt{2}+9}$
=$\sqrt{(\sqrt{2}+1)^{2}} +\sqrt{(3-\sqrt{2})² } =\sqrt{2}+1 +3-\sqrt{2}=4$
thay x=4 vào A ta có :
A=$\frac{\sqrt{4}(\sqrt{4}+2)}{4\sqrt{4}+1} =\frac{8}{9}$
