Đáp án:
c. x=4
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0\\
A = \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 4}}{{\sqrt x + 2}} = 1 + \dfrac{4}{{\sqrt x + 2}}\\
A \in {Z^ + } \Leftrightarrow \dfrac{4}{{\sqrt x + 2}} \in {Z^ + }\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 2 = 4\\
\sqrt x + 2 = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
b.A < \dfrac{{15}}{2}\\
\to \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} < \dfrac{{15}}{2}\\
\to \dfrac{{2\sqrt x + 12 - 15\sqrt x - 30}}{{2\left( {\sqrt x + 2} \right)}} < 0\\
\to \dfrac{{ - 13\sqrt x - 18}}{{2\left( {\sqrt x + 2} \right)}} < 0\\
\to \dfrac{{13\sqrt x + 18}}{{2\left( {\sqrt x + 2} \right)}} > 0\left( {ld} \right)\forall x \ge 0\\
KL:\forall x \ge 0\\
c.A = 2\\
\to \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} = 2\\
\to \sqrt x + 6 = 2\sqrt x + 4\\
\to \sqrt x = 2\\
\to x = 4
\end{array}\)
