Đáp án: 8) \(0 \le x \le \dfrac{1}{4}\) Giải thích các bước giải: \(\begin{array}{l}8)\\a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\P = \left[ {\dfrac{{\sqrt x + 2 + \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x + 2 – \sqrt x }}{{\sqrt x + 1}}} \right)\\ = \dfrac{{\sqrt x + 2 + x – 9 – x + 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\ = \dfrac{{\sqrt x – 3}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\ = \dfrac{{\sqrt x + 1}}{{x – 4}}\\b.\dfrac{1}{P} \le – \dfrac{5}{2}\\ \to \dfrac{{x – 4}}{{\sqrt x + 1}} \le – \dfrac{5}{2}\\ \to \dfrac{{2x – 8 + 5\sqrt x + 5}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\ \to \dfrac{{2x + 5\sqrt x – 3}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\ \to \dfrac{{\left( {2\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\ \to 2\sqrt x – 1 \le 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0;\sqrt x + 1\forall x \ge 0} \right)\\ \to \sqrt x \le \dfrac{1}{2}\\ \to x \le \dfrac{1}{4}\\ \to 0 \le x \le \dfrac{1}{4}\end{array}\) Reply
Đáp án:
8) \(0 \le x \le \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
8)\\
a.DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
P = \left[ {\dfrac{{\sqrt x + 2 + \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x + 2 – \sqrt x }}{{\sqrt x + 1}}} \right)\\
= \dfrac{{\sqrt x + 2 + x – 9 – x + 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x – 3}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 1}}{{x – 4}}\\
b.\dfrac{1}{P} \le – \dfrac{5}{2}\\
\to \dfrac{{x – 4}}{{\sqrt x + 1}} \le – \dfrac{5}{2}\\
\to \dfrac{{2x – 8 + 5\sqrt x + 5}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
\to \dfrac{{2x + 5\sqrt x – 3}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
\to \dfrac{{\left( {2\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
\to 2\sqrt x – 1 \le 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0;\sqrt x + 1\forall x \ge 0} \right)\\
\to \sqrt x \le \dfrac{1}{2}\\
\to x \le \dfrac{1}{4}\\
\to 0 \le x \le \dfrac{1}{4}
\end{array}\)