Đáp án:
c) \(\left[ \begin{array}{l}
m > 2\\
m < - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{2\sqrt x }}{{\sqrt x + m}} + \dfrac{{\sqrt x }}{{\sqrt x - m}} - \dfrac{{{m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \dfrac{{8\sqrt x \left( {\sqrt x - m} \right) + 4\sqrt x \left( {\sqrt x + m} \right) - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \dfrac{{8x - 8m\sqrt x + 4x + 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \dfrac{{12x - 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
b)P = 0\\
\to \dfrac{{12x - 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}} = 0\\
\to 12x - 4m\sqrt x - {m^2} = 0\\
Xét:\Delta ' = 4{m^2} - 12.\left( { - {m^2}} \right)\\
= 16{m^2}\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{2m + \sqrt {16m} }}{{12}}\\
\sqrt x = \dfrac{{2m - \sqrt {16m} }}{{12}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{m}{2}\\
\sqrt x = - \dfrac{m}{6}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{{m^2}}}{4}\\
c)x > 1\\
\to \dfrac{{{m^2}}}{4} > 1\\
\to {m^2} - 4 > 0\\
\to \left( {m - 2} \right)\left( {m + 2} \right) > 0\\
\to \left[ \begin{array}{l}
m > 2\\
m < - 2
\end{array} \right.
\end{array}\)
