Đáp án:
d. \(Min = - \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
A = \dfrac{{\sqrt x - 3 - 5 + \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 8 + x - 4}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + \sqrt x - 12}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} = \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}\\
DK:x \ge 0;x \ne 9\\
B = \dfrac{{{{\left( {\sqrt x + 4} \right)}^2}}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 4} \right)}} = \dfrac{{\sqrt x + 4}}{{\sqrt x - 3}}\\
b.Thay:x = \sqrt {\dfrac{6}{{1,5}}} = 2\\
\to B = \dfrac{{\sqrt 2 + 4}}{{\sqrt 2 + 2}} = 3 - \sqrt 2 \\
c.Q = \dfrac{A}{B} = \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}:\dfrac{{\sqrt x + 4}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 4}} = \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}}\\
d.Q = \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 5}}{{\sqrt x + 2}} = 1 - \dfrac{5}{{\sqrt x + 2}}\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{5}{{\sqrt x + 2}} \le \dfrac{5}{2}\\
\to - \dfrac{5}{{\sqrt x + 2}} \ge - \dfrac{5}{2}\\
\to 1 - \dfrac{5}{{\sqrt x + 2}} \ge - \dfrac{3}{2}\\
\to Min = - \dfrac{3}{2}\\
\Leftrightarrow x = 0\\
e.Q \in Z\\
\Leftrightarrow \dfrac{5}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 \in U\left( 5 \right)\\
Mà:\sqrt x + 2 \ge 2\forall x \ge 0\\
\to \sqrt x + 2 = 5\\
\to \sqrt x = 3\\
\to x = 9\left( l \right)\\
\to x \in \emptyset
\end{array}\)
