Đáp án:
c) \(\left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
b)A = \dfrac{{3x + 3\sqrt x - 3 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 3\sqrt x - 3 - x + 1 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
c)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}}\\
= 1 + \dfrac{2}{{\sqrt x - 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 2\\
\sqrt x - 1 = - 2\left( l \right)\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
