Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{4x}}{{\sqrt x - 3}}\\
b,\\
x = \dfrac{9}{{16}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\
= \left( {\dfrac{{4\sqrt x }}{{\sqrt x + 2}} - \dfrac{{8x}}{{x - 4}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\
= \left( {\dfrac{{4\sqrt x }}{{\sqrt x + 2}} - \dfrac{{8x}}{{{{\sqrt x }^2} - {2^2}}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \dfrac{2}{{\sqrt x }}} \right)\\
= \left( {\dfrac{{4\sqrt x }}{{\sqrt x + 2}} - \dfrac{{8x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\dfrac{{\left( {\sqrt x - 1} \right) - 2.\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4\sqrt x \left( {\sqrt x - 2} \right) - 8x}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\dfrac{{\sqrt x - 1 - 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4x - 8\sqrt x - 8x}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\dfrac{{ - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\,\,\,\,\,\,\,\,\,\,\,\\
= \dfrac{{ - 4x - 8\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\dfrac{{ - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 4\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x .\left( {\sqrt x - 2} \right)}}{{ - \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{4{{\sqrt x }^2}}}{{\sqrt x - 3}}\\
= \dfrac{{4x}}{{\sqrt x - 3}}\\
b,\\
P = - 1\\
\Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} = - 1\\
\Leftrightarrow 4x = - \left( {\sqrt x - 3} \right)\\
\Leftrightarrow 4x + \sqrt x - 3 = 0\\
\Leftrightarrow \left( {4x + 4\sqrt x } \right) + \left( { - 3\sqrt x - 3} \right) = 0\\
\Leftrightarrow 4\sqrt x \left( {\sqrt x + 1} \right) - 3.\left( {\sqrt x + 1} \right) = 0\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( {4\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 0\\
4\sqrt x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 1\\
\sqrt x = \dfrac{3}{4}
\end{array} \right.\\
\sqrt x > 0,\,\,\,\forall x > 0,x \ne 4,x \ne 9\\
\Rightarrow \sqrt x = \dfrac{3}{4}\\
\Leftrightarrow x = \dfrac{9}{{16}}
\end{array}\)
