Đáp án:
c. \(0 \le x < 4\)
Giải thích các bước giải:
\(\begin{array}{l}
a.P = \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}\\
b.Thay:x = 9 + 4\sqrt 2 \\
= 8 + 2.2\sqrt 2 .1 + 1\\
= {\left( {2\sqrt 2 + 1} \right)^2}\\
\to Q = \dfrac{{\sqrt x + 3}}{{\sqrt x + 1}} = \dfrac{{\sqrt {{{\left( {2\sqrt 2 + 1} \right)}^2}} + 3}}{{\sqrt {{{\left( {2\sqrt 2 + 1} \right)}^2}} + 1}}\\
= \dfrac{{2\sqrt 2 + 1 + 3}}{{2\sqrt 2 + 1 + 1}} = \dfrac{{2\sqrt 2 + 4}}{{2\sqrt 2 + 2}} = \sqrt 2 \\
c.M = P.Q = \dfrac{3}{{\sqrt x + 3}}.\dfrac{{\sqrt x + 3}}{{\sqrt x + 1}}\\
= \dfrac{3}{{\sqrt x + 1}}\\
M > 1\\
\to \dfrac{3}{{\sqrt x + 1}} > 1\\
\to \dfrac{{3 - \sqrt x - 1}}{{\sqrt x + 1}} > 0\\
\to \dfrac{{2 - \sqrt x }}{{\sqrt x + 1}} > 0\\
\to 2 - \sqrt x > 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0;x \ne 9} \right)\\
\to 4 > x\\
\to 0 \le x < 4\\
d.M = \dfrac{3}{{\sqrt x + 1}}\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{3}{{\sqrt x + 1}} \le 3\\
\to Max = 3\\
\Leftrightarrow x = 0
\end{array}\)
