Đáp án:
$A =\dfrac{2}{\sqrt[]{x}-1}$ với $x≥0$
Giải thích các bước giải:
$ĐKXĐ : x ≥ 0 $
Ta có : $A = \bigg(\dfrac{2+\sqrt[]{x}}{x+2\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}-2}{x-1}\bigg).\dfrac{x+2\sqrt[]{x}+1}{\sqrt[]{x}}$
$ = \bigg[\dfrac{2+\sqrt[]{x}}{(\sqrt[]{x}+1)^2} - \dfrac{\sqrt[]{x}-2}{x-1}\bigg].\dfrac{x+2\sqrt[]{x}+1}{\sqrt[]{x}}$
$ = \bigg[\dfrac{(2+\sqrt[]{x}).(x-1)-(\sqrt[]{x}-2).(\sqrt[]{x}+1)^2}{(\sqrt[]{x}+1)^2.(x-1)}\bigg].\dfrac{(\sqrt[]{x}+1)^2}{\sqrt[]{x}}$
$ = \dfrac{2\sqrt[]{x}.(\sqrt[]{x}+1)}{(\sqrt[]{x}+1)^2.(x-1)}.\dfrac{(\sqrt[]{x}+1)^2}{\sqrt[]{x}}$
$ = \dfrac{2\sqrt[]{x}}{(\sqrt[]{x}+1).(x-1)}.\dfrac{(\sqrt[]{x}+1)^2}{\sqrt[]{x}}$
$ = \dfrac{2.(\sqrt[]{x}+1)}{x-1}$
$ = \dfrac{2.(\sqrt[]{x}+1)}{(\sqrt[]{x}+1).(\sqrt[]{x}-1)}$
$ = \dfrac{2}{\sqrt[]{x}-1}$
Vậy $A =\dfrac{2}{\sqrt[]{x}-1} $ với $x≥0$
