Đáp án:
Bài 2:
a.$a\ge 0, a\ne 4$
Giải thích các bước giải:
Bài 2:
$Q=(\dfrac{1}{\sqrt[]{a}-2}-\dfrac{1}{\sqrt[]{a}}):(\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}+1}-\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2})$
$=\dfrac{\sqrt[]{a}-(\sqrt[]{a}-2)}{\sqrt[]{a}.(\sqrt[]{a}-2)}:\dfrac{(\sqrt[]{a}-2)(\sqrt[]{a}+2)-(\sqrt[]{a}+1)^2}{(\sqrt[]{a}+1)(\sqrt[]{a}-2)}$
$=\dfrac{2}{\sqrt[]{a}(\sqrt[]{a}-2)}:\dfrac{a-4-(a+2\sqrt[]{a}+1)}{(\sqrt[]{a}+1)(\sqrt[]{a}-2)}$
$=\dfrac{2}{\sqrt[]{a}(\sqrt[]{a}-2)}:\dfrac{-2\sqrt[]{a}-5}{(\sqrt[]{a}+1)(\sqrt[]{a}-2)}$
a.DKXD:
$\begin{cases}a\ge 0\\\sqrt[]{a}-2\ne 0\\-2\sqrt[]{a}-5\ne 0\\\sqrt[]{a}+1\ne 0\\\sqrt[]{a}-2\ne 0\end{cases}\rightarrow \begin{cases}a\ge 0\\ a\ne 4\end{cases}$
b.Ta có:
$Q=\dfrac{2}{\sqrt[]{a}(\sqrt[]{a}-2)}:\dfrac{-2\sqrt[]{a}-5}{(\sqrt[]{a}+1)(\sqrt[]{a}-2)}$
$\rightarrow Q=\dfrac{2}{\sqrt[]{a}(\sqrt[]{a}-2)}.\dfrac{(\sqrt[]{a}+1)(\sqrt[]{a}-2)}{-2\sqrt[]{a}-5}$
$\rightarrow Q=\dfrac{-2(\sqrt[]{a}+1)}{\sqrt[]{a}(2\sqrt[]{a}+5)}$
$\rightarrow Q>0\leftrightarrow \dfrac{-2(\sqrt[]{a}+1)}{\sqrt[]{a}(2\sqrt[]{a}+5)}>0$
Mà $\dfrac{-2(\sqrt[]{a}+1)}{\sqrt[]{a}(2\sqrt[]{a}+5)}<0\quad\forall a\in DKXD$
$\rightarrow $ không có a thỏa mãn đề
