Đáp án:
a) \(\dfrac{{\sqrt x }}{{\sqrt x + 3}}\)
Giải thích các bước giải:
Câu 1:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x + 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{x}{{\sqrt x \left( {\sqrt x - 3} \right)}} + \dfrac{{2\sqrt x + 5}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{{2\sqrt x + 5}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 15 - \sqrt x \left( {\sqrt x + 3} \right) + \left( {2\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 15 - x - 3\sqrt x + 2x - \sqrt x - 15}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
b)A = 2B\\
\to \dfrac{{\sqrt x }}{{\sqrt x + 3}} = 2.\dfrac{{8\sqrt x - 3}}{{14}}\\
\to \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{8\sqrt x - 3}}{7}\\
\to \left( {8\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) = 7\sqrt x \\
\to 8x + 21\sqrt x - 9 = 7\sqrt x \\
\to \left( {2\sqrt x - 1} \right)\left( {4\sqrt x + 9} \right) = 0\\
\to 2\sqrt x - 1 = 0\\
\to x = \dfrac{1}{4}\\
c)A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 - 3}}{{\sqrt x + 3}}\\
= 1 - \dfrac{3}{{\sqrt x + 3}}\\
A \in Z\\
\to \dfrac{3}{{\sqrt x + 3}} \in Z\\
\to \sqrt x + 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 3 = 3\\
\sqrt x + 3 = 1\left( l \right)
\end{array} \right.\\
\to x = 0\left( {KTM} \right)\\
\to dpcm
\end{array}\)
