Giải giúp mk bài trong hình vs. Question Giải giúp mk bài trong hình vs. in progress 0 Môn Toán Delwyn 4 years 2020-10-29T09:18:37+00:00 2020-10-29T09:18:37+00:00 1 Answers 47 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} – \dfrac{{\sqrt x – 1}}{{\sqrt x + 2}} + \dfrac{{4\sqrt x – 4}}{{4 – x}}} \right):\left( {1 + \dfrac{5}{{\sqrt x – 2}}} \right)\\
= \left( {\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} – \dfrac{{\sqrt x – 1}}{{\sqrt x + 2}} – \dfrac{{4\sqrt x – 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\dfrac{{\left( {\sqrt x – 2} \right) + 5}}{{\sqrt x – 2}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 2} \right) – \left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right) – \left( {4\sqrt x – 4} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}}\\
= \dfrac{{\left( {x + 5\sqrt x + 6} \right) – \left( {x – 3\sqrt x + 2} \right) – 4\sqrt x + 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}}\\
= \dfrac{{4\sqrt x + 8}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x – 2}}{{\sqrt x + 3}}\\
= \dfrac{{4.\left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}.\dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{4}{{\sqrt x + 3}}\\
b,\\
P > \dfrac{1}{2}\\
\Leftrightarrow \dfrac{4}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\Leftrightarrow 8 > \sqrt x + 3\\
\Leftrightarrow \sqrt x < 5\\
\Leftrightarrow \left\{ \begin{array}{l}
0 \le x < 25\\
x \ne 4
\end{array} \right.\\
c,\\
P.Q = \dfrac{4}{{\sqrt x + 3}}.\dfrac{{3\left( {\sqrt x + 3} \right)}}{{4.\left( {\sqrt x + 1} \right)}} = \dfrac{3}{{\sqrt x + 1}}\\
\sqrt x + 1 \ge 1,\,\,\,\forall x \ge 0,x \ne 4 \Rightarrow \dfrac{3}{{\sqrt x + 1}} \le \dfrac{3}{1} = 3 \Rightarrow P.Q \le 3\\
\sqrt x + 1 > 0 \Rightarrow \dfrac{3}{{\sqrt x + 1}} > 0 \Rightarrow P.Q > 0\\
\Rightarrow 0 < P.Q \le 3\\
P.Q \in Z \Rightarrow P.Q = \left\{ {1;2;3} \right\}\\
P.Q = 1 \Leftrightarrow \dfrac{3}{{\sqrt x + 1}} = 1 \Leftrightarrow \sqrt x + 1 = 3 \Leftrightarrow x = 4\,\,\,\,\left( {L,\,\,\,x \ne 4} \right)\\
P.Q = 2 \Leftrightarrow \dfrac{3}{{\sqrt x + 1}} = 2 \Leftrightarrow \sqrt x + 1 = \dfrac{3}{2} \Leftrightarrow \sqrt x = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}\,\,\,\,\left( {t/m} \right)\\
P.Q = 3 \Leftrightarrow \dfrac{3}{{\sqrt x + 1}} = 3 \Leftrightarrow \sqrt x + 1 = 1 \Leftrightarrow x = 0\,\,\,\,\left( {t/m} \right)
\end{array}\)