Đáp án:
b) \(\dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:x = 16\\
\to A = \dfrac{{\sqrt {16} - 1}}{{2\sqrt {16} + 10}} = \dfrac{{4 - 1}}{{2.4 + 10}} = \dfrac{3}{{18}} = \dfrac{1}{6}\\
b)B = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) + 2\left( {\sqrt x - 2} \right) - 9\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 4\sqrt x + 3 + 2\sqrt x - 4 - 9\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\\
c)P = B:A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}:\dfrac{{\sqrt x - 1}}{{2\sqrt x + 10}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}.\dfrac{{2\sqrt x + 10}}{{\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 10}}{{\sqrt x + 3}} = \dfrac{{2\left( {\sqrt x + 3} \right) + 4}}{{\sqrt x + 3}}\\
= 2 + \dfrac{4}{{\sqrt x + 3}}\\
P \in Z\\
\Leftrightarrow \dfrac{4}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 \in U\left( 4 \right)\\
Mà:\sqrt x + 3 \ge 3\forall x \ge 0\\
\to \sqrt x + 3 = 4\\
\to \sqrt x = 1\\
\to x = 1\left( l \right)\\
\to x \in \emptyset
\end{array}\)
