mấy anh chị giúp em vs ạ

Đáp án:

a) $\left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$

b) $A=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$

c) $A=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$

Giải thích các bước giải:

a) ĐKXĐ: 

$\left\{ \begin{array}{l}\sqrt[]{a}-1 \neq 0\\\sqrt[]{a} \neq 0\\\sqrt[]{a}-2 \neq 0\\a≥0\end{array} \right.$

$↔ \left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$

b) $A=\Bigg(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\Bigg):\Bigg(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\Bigg)$

$=\dfrac{\sqrt[]{a}-(\sqrt[]{a}-1)}{\sqrt[]{a}(\sqrt[]{a}-1)}:\dfrac{(\sqrt[]{a}+1)(\sqrt[]{a}-1)-(\sqrt[]{a}+2)(\sqrt[]{a}-2)}{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}$

$=\dfrac{1}{\sqrt[]{a}(\sqrt[]{a}-1)}.\dfrac{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}{a-1-a+4}$

$=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$

c) $A=\dfrac{1}{9} ↔ \dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}=\dfrac{1}{9}$

$↔ 3\sqrt[]{a}=9\sqrt[]{a}-18$

$↔ 6\sqrt[]{a}=18$

$↔ \sqrt[]{a}=3$

$→ a=9$

d) $a=9+4\sqrt[]{5}$

$→ \sqrt[]{a}=\sqrt[]{9+4\sqrt[]{5}}$

$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5})^2+2.\sqrt[]{5}.2+4}$

$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5}+2)^2}$

$↔ \sqrt[]{a}=|\sqrt[]{5}+2|$

$↔ \sqrt[]{a}=\sqrt[]{5}+2$

Vậy $A=\dfrac{\sqrt[]{5}+2-2}{3(\sqrt[]{5}+2)}$

$=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$