Question

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Answer 1

\[\begin{array}{l}
P = \frac{{2\sqrt x }}{{\sqrt x + m}} + \frac{{\sqrt x }}{{\sqrt x - m}} - \frac{{{m^2}}}{{4x - 4{m^2}}}\;\\
dk:m\; > 0;x \ne {m^2}\\
a)\\
P = \frac{{2\sqrt x }}{{\sqrt x + m}} + \frac{{\sqrt x }}{{\sqrt x - m}} - \frac{{{m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{8\sqrt x \left( {\sqrt x - m} \right) + 4\sqrt x \left( {\sqrt x + m} \right) - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{8x - 8m\sqrt x + 4x + 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{12x - 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{12x - 6m\sqrt x + 2m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{6\sqrt x \left( {2\sqrt x - m} \right) + m\left( {2\sqrt x - m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{\left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
b)P = 0\left( {m > 0;x \ne {m^2}} \right)\\
\Leftrightarrow \frac{{\left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}} = 0\\
\Leftrightarrow \left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\sqrt x - m = 0\\
6\sqrt x + m = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \frac{m}{2}\\
\sqrt x = \frac{{ - m}}{6}\left( {ktm\_do:m > 0} \right)
\end{array} \right.\\
\Leftrightarrow x = \frac{{{m^2}}}{4}\left( {tm} \right)
\end{array}\]

vậy với \[x = \frac{{{m^2}}}{4}\] thì P=0

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Answer 2

\[\begin{array}{l}
P = \frac{{2\sqrt x }}{{\sqrt x + m}} + \frac{{\sqrt x }}{{\sqrt x - m}} - \frac{{{m^2}}}{{4x - 4{m^2}}}\;\\
dk:m\; > 0;x \ne {m^2}\\
a)\\
P = \frac{{2\sqrt x }}{{\sqrt x + m}} + \frac{{\sqrt x }}{{\sqrt x - m}} - \frac{{{m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{8\sqrt x \left( {\sqrt x - m} \right) + 4\sqrt x \left( {\sqrt x + m} \right) - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{8x - 8m\sqrt x + 4x + 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{12x - 4m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{12x - 6m\sqrt x + 2m\sqrt x - {m^2}}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{6\sqrt x \left( {2\sqrt x - m} \right) + m\left( {2\sqrt x - m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
= \frac{{\left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}}\\
b)P = 0\left( {m > 0;x \ne {m^2}} \right)\\
\Leftrightarrow \frac{{\left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right)}}{{4\left( {\sqrt x - m} \right)\left( {\sqrt x + m} \right)}} = 0\\
\Leftrightarrow \left( {2\sqrt x - m} \right)\left( {6\sqrt x + m} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\sqrt x - m = 0\\
6\sqrt x + m = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \frac{m}{2}\\
\sqrt x = \frac{{ - m}}{6}\left( {ktm\_do:m > 0} \right)
\end{array} \right.\\
\Leftrightarrow x = \frac{{{m^2}}}{4}\left( {tm} \right)
\end{array}\]

vậy với \[x = \frac{{{m^2}}}{4}\] thì P=0

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