Đáp án:
Giải thích các bước giải:
đkxđ: x ≥0; x#1/25
$\begin{array}{l}
1)Q = \frac{{ - 7(\sqrt x + 7)}}{{5\sqrt x - 1}} + \frac{{2\sqrt x - 2}}{{\sqrt x + 2}} + \frac{{39\sqrt x + 12}}{{5x + 9\sqrt x - 2}}\\
= \frac{{ - 7(\sqrt x + 7)(\sqrt x + 2)}}{{(5\sqrt x - 1)(\sqrt x + 2)}} + \frac{{(2\sqrt x - 2)(5\sqrt x - 1)}}{{(\sqrt x + 2)(5\sqrt x - 1)}} + \frac{{39\sqrt x + 12}}{{(\sqrt x + 2)(5\sqrt x - 1)}}\\
= \frac{{ - 7x - 63\sqrt x - 98 + 10x - 12\sqrt x + 2 + 39\sqrt x + 12}}{{(\sqrt x + 2)(5\sqrt x - 1)}}\\
= \frac{{3x - 36\sqrt x - 84}}{{(\sqrt x + 2)(5\sqrt x - 1)}} = \frac{{(\sqrt x + 2)(\sqrt x - 14)}}{{(\sqrt x + 2)(5\sqrt x - 1)}} = \frac{{\sqrt x - 14}}{{5\sqrt x - 1}}\\
2)Q \le - 3\\
\Leftrightarrow \frac{{\sqrt x - 14}}{{5\sqrt x - 1}} \le - 3\\
\Leftrightarrow \frac{{\sqrt x - 14}}{{5\sqrt x - 1}} + 3 \le 0\\
\Leftrightarrow \frac{{16\sqrt x - 17}}{{5\sqrt x - 1}} \le 0\\
\Leftrightarrow a)16\sqrt x - 17 \le 0\,va\,5\sqrt x - 1 \ge 0\\
\Leftrightarrow x \le \frac{{289}}{{256}}\,va\,x \ge \frac{1}{{25}}\,hay\,\frac{1}{{25}} \le x \le \frac{{289}}{{256}}\,\\
b)16\sqrt x - 17 \ge 0\,va\,5\sqrt x - 1 \le 0\\
ko\,tim\,dc\,x\,trong\,THb\\
\,vay\frac{1}{{25}} < x \le \frac{{289}}{{256}}\,thi\,Q \le - 3
\end{array}$
