Đáp án:
c. \(x = \dfrac{9}{{25}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 4\\
B = \left( {\dfrac{{2 + \sqrt x }}{{2 - \sqrt x }} - \dfrac{{2 - \sqrt x }}{{2 + \sqrt x }} - \dfrac{{4x}}{{x - 4}}} \right):\left( {\dfrac{2}{{2 - \sqrt x }} - \dfrac{{\sqrt x + 3}}{{2\sqrt x - x}}} \right)\\
= \left[ {\dfrac{{{{\left( {2 + \sqrt x } \right)}^2} - {{\left( {2 - \sqrt x } \right)}^2} + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}} \right].\left[ {\dfrac{{2\sqrt x - \sqrt x - 3}}{{\sqrt x \left( {2 - \sqrt x } \right)}}} \right]\\
= \dfrac{{4 + 4\sqrt x + x - 4 + 4\sqrt x - x + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\\
= \dfrac{{4x + 8\sqrt x }}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4\left( {2 + \sqrt x } \right)}}{{2 + \sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{4\sqrt x }}{{\sqrt x - 3}}\\
b.B > 0\\
\to \dfrac{{4\sqrt x }}{{\sqrt x - 3}} > 0\\
\to \sqrt x - 3 > 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to x > 9\\
c.B = - 1\\
\to \dfrac{{4\sqrt x }}{{\sqrt x - 3}} = - 1\\
\to 4\sqrt x = - \sqrt x + 3\\
\to 5\sqrt x = 3\\
\to \sqrt x = \dfrac{3}{5}\\
\to x = \dfrac{9}{{25}}\left( {TM} \right)
\end{array}\)
