Đáp án:
c) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 4\\
P = \dfrac{{x + 8}}{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}} - \dfrac{1}{{\sqrt x + 2}}\\
= \dfrac{{x + 8 - x + 2\sqrt x - 4}}{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}}\\
= \dfrac{{2\sqrt x - 4}}{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}} = \dfrac{{2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}}\\
b)A = P.Q = \dfrac{{2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}}.\dfrac{{x - 2\sqrt x + 4}}{{\sqrt x - 2}}\\
= \dfrac{2}{{\sqrt x + 2}}\\
c)A \in Z\\
\to \dfrac{2}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 2\\
\sqrt x + 2 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = - 1\left( l \right)
\end{array} \right.\\
\to x = 0
\end{array}\)
