Đáp án:
\(a.Q = \dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne \dfrac{4}{9}\\
Q = \dfrac{{\left( {3 - \sqrt x } \right)\left( {3\sqrt x - 2} \right) + \left( {3\sqrt x + 4} \right)\left( {5\sqrt x + 7} \right) - 42\sqrt x - 34}}{{\left( {3\sqrt x - 2} \right)\left( {5\sqrt x + 7} \right)}}\\
= \dfrac{{9\sqrt x - 6 - 3x + 2\sqrt x + 15x + 21\sqrt x + 20\sqrt x + 28 - 42\sqrt x - 34}}{{\left( {3\sqrt x - 2} \right)\left( {5\sqrt x + 7} \right)}}\\
= \dfrac{{12x + 10\sqrt x - 12}}{{\left( {3\sqrt x - 2} \right)\left( {5\sqrt x + 7} \right)}}\\
= \dfrac{{\left( {3\sqrt x - 2} \right)\left( {2\sqrt x + 3} \right)}}{{\left( {3\sqrt x - 2} \right)\left( {5\sqrt x + 7} \right)}}\\
= \dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}}\\
b)Q \le - \dfrac{2}{3}\\
\to \dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}} \le - \dfrac{2}{3}\\
\to \dfrac{{6\sqrt x + 9 + 10\sqrt x + 14}}{{3\left( {5\sqrt x + 7} \right)}} \le 0\\
\to \dfrac{{16\sqrt x + 23}}{{3\left( {5\sqrt x + 7} \right)}} \le 0\\
Do:x \ge 0 \to \left\{ \begin{array}{l}
16\sqrt x + 23 > 0\\
5\sqrt x + 7 > 0
\end{array} \right.\\
\to \dfrac{{16\sqrt x + 23}}{{3\left( {5\sqrt x + 7} \right)}} > 0\\
\to \dfrac{{16\sqrt x + 23}}{{3\left( {5\sqrt x + 7} \right)}} \le 0\left( {vô lý} \right)
\end{array}\)
⇒ Không tồn tại giá trị x TMĐK
\(\begin{array}{l}
c.Q = \dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}}\\
\to 5Q = \dfrac{{10\sqrt x + 15}}{{5\sqrt x + 7}} = \dfrac{{2\left( {5\sqrt x + 7} \right) + 1}}{{5\sqrt x + 7}}\\
= 2 + \dfrac{1}{{5\sqrt x + 7}}\\
Q \in Z\\
\Leftrightarrow \dfrac{1}{{5\sqrt x + 7}} \in Z\\
\Leftrightarrow 5\sqrt x + 7 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
5\sqrt x + 7 = 1\\
5\sqrt x + 7 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
5\sqrt x = - 6\left( l \right)\\
5\sqrt x = - 8\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
