Đáp án:
\(\begin{array}{l}
a)\dfrac{{3\sqrt a }}{{\sqrt a - 2}}\\
b)a \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Q = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right) - \left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}:\dfrac{{\sqrt a - 1 - \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}\\
= \dfrac{{a - 4 - a + 1}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{ - 1}}\\
= \dfrac{{3\sqrt a }}{{\sqrt a - 2}}\\
b)\left| Q \right| + Q = 0\\
\to \left[ \begin{array}{l}
Q + Q = 0\\
- Q + Q = 0\left( {ld} \right)
\end{array} \right.\\
\to Q = 0\\
\to \dfrac{{3\sqrt a }}{{\sqrt a - 2}} = 0\\
\to \sqrt a = 0\\
\to a = 0\left( {KTM} \right)\\
\to a \in \emptyset
\end{array}\)
