Đáp án:
c. \(4 < x < 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 4;x \ne 9\\
Q = \left[ {\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) - 2\sqrt x \left( {\sqrt x - 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}} \right]:\left[ {\frac{{\sqrt x \left( {3 - \sqrt x } \right)}}{{{{\left( {\sqrt x + 2} \right)}^2}}}} \right]\\
= \frac{{x + 3\sqrt x + 2 - 2x + 4\sqrt x - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\frac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\
= \frac{{ - x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\frac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\
= \frac{{ - \sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\frac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\
= \frac{{ - {{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {3 - \sqrt x } \right)}}\\
b.Q = 2\\
\to \frac{{ - {{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {3 - \sqrt x } \right)}} = 2\\
\to - x - 4\sqrt x - 4 = 2\left( { - x + 5\sqrt x - 6} \right)\\
\to - x - 4\sqrt x - 4 = - 2x + 10\sqrt x - 12\\
\to x - 14\sqrt x + 8 = 0\\
Có: Δ'= {7^2} - 8 = 41 > 0\\
\to \left[ \begin{array}{l}
\sqrt x = 7 + \sqrt {41} \\
\sqrt x = 7 - \sqrt {41}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 90 + 14\sqrt {41} \\
x = 90 - 14\sqrt {41}
\end{array} \right.\\
c.Q < 0\\
\to \frac{{ - {{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {3 - \sqrt x } \right)}} < 0\\
\to \frac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {3 - \sqrt x } \right)}} > 0\\
Do:{\left( {\sqrt x + 2} \right)^2} > 0\forall x \ge 0\\
\to \left( {\sqrt x - 2} \right)\left( {3 - \sqrt x } \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x - 2 > 0\\
3 - \sqrt x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt x - 2 < 0\\
3 - \sqrt x < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 4\\
9 > x
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 4\\
9 < x
\end{array} \right.\left( l \right)
\end{array} \right.\\
KL:4 < x < 9
\end{array}\)
