Question

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Answer 1

\[\begin{array}{l}
P = \frac{{3\left( {x + \sqrt x - 3} \right)}}{{x + \sqrt x - 2}} + \frac{{\sqrt x + 3}}{{\sqrt x + 2}} - \frac{{\sqrt x - 2}}{{\sqrt x - 1}}\;\\
dk:x \ge 0\;,\;x \ne 1\\
a)P = \frac{{3\left( {x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} - \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3x + 3\sqrt x - 9 + x - \sqrt x + 3\sqrt x - 3 - \left( {x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{4x + 5\sqrt x - 12 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3x + 5\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 8} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
b)P < \frac{{15}}{4}\\
\Leftrightarrow \frac{{3\sqrt x + 8}}{{\sqrt x + 2}} < \frac{{15}}{4}\left( {x \ge 0\;,\;x \ne 1} \right)\\
\Leftrightarrow \frac{{3\sqrt x + 8}}{{\sqrt x + 2}} - \frac{{15}}{4} < 0\\
\Leftrightarrow \frac{{12\sqrt x + 32 - 15\sqrt x - 30}}{{4\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow \frac{{ - 3\sqrt x + 2}}{{4\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow - 3\sqrt x + 2 < 0\left( {do:4\left( {\sqrt x + 2} \right) > 0} \right)\\
\Leftrightarrow \sqrt x > \frac{2}{3}\\
\Leftrightarrow x > \frac{4}{9}
\end{array}\]

vậy với \[x > \frac{4}{9};x \ne 1\] thì \[P < \frac{{15}}{4}\]

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Answer 2

\[\begin{array}{l}
P = \frac{{3\left( {x + \sqrt x - 3} \right)}}{{x + \sqrt x - 2}} + \frac{{\sqrt x + 3}}{{\sqrt x + 2}} - \frac{{\sqrt x - 2}}{{\sqrt x - 1}}\;\\
dk:x \ge 0\;,\;x \ne 1\\
a)P = \frac{{3\left( {x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} - \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3x + 3\sqrt x - 9 + x - \sqrt x + 3\sqrt x - 3 - \left( {x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{4x + 5\sqrt x - 12 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3x + 5\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 8} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
b)P < \frac{{15}}{4}\\
\Leftrightarrow \frac{{3\sqrt x + 8}}{{\sqrt x + 2}} < \frac{{15}}{4}\left( {x \ge 0\;,\;x \ne 1} \right)\\
\Leftrightarrow \frac{{3\sqrt x + 8}}{{\sqrt x + 2}} - \frac{{15}}{4} < 0\\
\Leftrightarrow \frac{{12\sqrt x + 32 - 15\sqrt x - 30}}{{4\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow \frac{{ - 3\sqrt x + 2}}{{4\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow - 3\sqrt x + 2 < 0\left( {do:4\left( {\sqrt x + 2} \right) > 0} \right)\\
\Leftrightarrow \sqrt x > \frac{2}{3}\\
\Leftrightarrow x > \frac{4}{9}
\end{array}\]

vậy với \[x > \frac{4}{9};x \ne 1\] thì \[P < \frac{{15}}{4}\]

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