Đáp án:
a) \(\dfrac{{\sqrt x }}{{\sqrt x - 3}}\)
b) \(\left[ \begin{array}{l}
x = 36\\
x = 16\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 - x - 4}}{{x + \sqrt x + 1}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
b)A = \dfrac{{\sqrt x }}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 3}}{{\sqrt x - 3}}\\
= 1 + \dfrac{3}{{\sqrt x - 3}}\\
A \in Z \to \dfrac{3}{{\sqrt x - 3}} \in Z\\
\to \sqrt x - 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 3\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1\\
\sqrt x - 3 = - 3
\end{array} \right. \to \left[ \begin{array}{l}
x = 36\\
x = 16\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
