Question

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Answer 1

\[\begin{array}{l}
P = \left( {\sqrt x - \frac{{x + 2}}{{\sqrt x + 1}}} \right):\left( {\frac{{\sqrt x }}{{\sqrt x + 1}} - \frac{{\sqrt x - 4}}{{1 - x}}} \right)\\
a)dkxd:\\
\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x + 1 \ne 0\\
1 - x \ne 0\\
x - 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 4
\end{array} \right.\\
b)P = \frac{{\sqrt x \left( {\sqrt x + 1} \right) - x - 2}}{{\sqrt x + 1}}:\left[ {\frac{{\sqrt x }}{{\sqrt x + 1}} + \frac{{\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \frac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}:\frac{{\sqrt x \left( {\sqrt x - 1} \right) + \sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x - 2}}{{\sqrt x + 1}}.\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x - \sqrt x + \sqrt x - 4}}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{x - 4}}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
c)P < 0\\
= > \frac{{\sqrt x - 1}}{{\sqrt x + 2}} < 0\left( {x \ge 0;x \ne 1;x \ne 4} \right)\\
\Leftrightarrow \sqrt x - 1 < 0\left( {do:\sqrt x + 2 > 0} \right)\\
= > \sqrt x < 1\\
= > x < 1
\end{array}\]

Vậy với \[0 \le x < 1\] thì P<0

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Answer 2

\[\begin{array}{l}
P = \left( {\sqrt x - \frac{{x + 2}}{{\sqrt x + 1}}} \right):\left( {\frac{{\sqrt x }}{{\sqrt x + 1}} - \frac{{\sqrt x - 4}}{{1 - x}}} \right)\\
a)dkxd:\\
\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x + 1 \ne 0\\
1 - x \ne 0\\
x - 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 4
\end{array} \right.\\
b)P = \frac{{\sqrt x \left( {\sqrt x + 1} \right) - x - 2}}{{\sqrt x + 1}}:\left[ {\frac{{\sqrt x }}{{\sqrt x + 1}} + \frac{{\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \frac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}:\frac{{\sqrt x \left( {\sqrt x - 1} \right) + \sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x - 2}}{{\sqrt x + 1}}.\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x - \sqrt x + \sqrt x - 4}}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{x - 4}}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
c)P < 0\\
= > \frac{{\sqrt x - 1}}{{\sqrt x + 2}} < 0\left( {x \ge 0;x \ne 1;x \ne 4} \right)\\
\Leftrightarrow \sqrt x - 1 < 0\left( {do:\sqrt x + 2 > 0} \right)\\
= > \sqrt x < 1\\
= > x < 1
\end{array}\]

Vậy với \[0 \le x < 1\] thì P<0

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