Đáp án:
c) \(36 > x \ge 0;x \ne 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 64\\
A = \dfrac{{\sqrt {64} + 1}}{{\sqrt {64} - 3}} = \dfrac{9}{5}\\
b)B = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
c)P = B:A = \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}\\
P < - \dfrac{1}{3}\\
\to - \dfrac{3}{{\sqrt x + 3}} < - \dfrac{1}{3}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{3}\\
\to \dfrac{{9 - \sqrt x - 3}}{{3\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{6 - \sqrt x }}{{3\left( {\sqrt x + 3} \right)}} > 0\\
\to 6 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0} \right)\forall x \ge 0\\
\to 36 > x \ge 0;x \ne 9
\end{array}\)
