Đáp án:
a. \(A = \frac{{\sqrt x }}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \frac{{\sqrt x + 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \frac{x}{{\sqrt x \left( {\sqrt x - 3} \right)}} + \frac{{2\sqrt x + 5}}{{\sqrt x + 3}}\\
= \frac{{\sqrt x + 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\sqrt x }}{{\sqrt x - 3}} + \frac{{2\sqrt x + 5}}{{\sqrt x + 3}}\\
= \frac{{\sqrt x + 15 - x - 3\sqrt x + 2x - \sqrt x - 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{x - 3\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\sqrt x }}{{\sqrt x + 3}}\\
b.A = 2B\\
\to \frac{{\sqrt x }}{{\sqrt x + 3}} = 2.\frac{{8\sqrt x - 3}}{{14}}\\
\to \frac{{\sqrt x }}{{\sqrt x + 3}} = \frac{{8\sqrt x - 3}}{7}\\
\to 7\sqrt x = 8x + 21\sqrt x - 9\\
\to 8x + 14\sqrt x - 9 = 0\\
\to 8x - 4\sqrt x + 18\sqrt x - 9 = 0\\
\to \left( {2\sqrt x - 1} \right)\left( {4\sqrt x + 9} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \frac{1}{2}\\
\sqrt x = - \frac{9}{4}\left( l \right)
\end{array} \right.\\
\to x = \frac{1}{4}\\
c.A = \frac{{\sqrt x }}{{\sqrt x + 3}} = \frac{{\sqrt x + 3 - 3}}{{\sqrt x + 3}} = 1 - \frac{3}{{\sqrt x + 3}}\\
Để:A \in Z\\
\Leftrightarrow \frac{3}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 \in U\left( 3 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 3 = 3\\
\sqrt x + 3 = - 3\left( l \right)\\
\sqrt x + 3 = 1\left( l \right)\\
\sqrt x + 3 = - 1\left( l \right)
\end{array} \right.\\
\to \sqrt x = 0\\
\to x = 0\left( l \right)
\end{array}\)
⇒ Không tồn tại giá trị x để A nguyên
