P = ( $\frac{2\sqrt{x}}{\sqrt{x}+3}$ + $\frac{\sqrt{x}}{\sqrt{x}-3}$ – $\frac{3(\sqrt{x}+3)}{x-9}$ ) : ( $\frac{2\sqrt{x}-2}{\sqrt{x}-3}$ -1)
a) rút gọn P
b) tìm x để P<-1
c) tìm x>4 nguyên để p có giá trị nguyên $\sqrt{x}$ = $\frac{\sqrt{3}+1}{\sqrt{2}}$
Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x – 3}} – \dfrac{{3\left( {\sqrt x + 3} \right)}}{{x – 9}}} \right):\left( {\dfrac{{2\sqrt x – 2}}{{\sqrt x – 3}} – 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x – 3}} – \dfrac{{3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\dfrac{{\left( {2\sqrt x – 2} \right) – \left( {\sqrt x – 3} \right)}}{{\sqrt x – 3}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) – 3.\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{2x – 6\sqrt x + x + 3\sqrt x – 3\sqrt x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\
= \dfrac{{3x – 6\sqrt x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{{3.\left( {x – 2\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{{3.\left( {\sqrt x + 1} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{{3.\left( {\sqrt x – 3} \right)}}{{\sqrt x + 3}}\\
b,\\
P < – 1 \Leftrightarrow \dfrac{{3.\left( {\sqrt x – 3} \right)}}{{\sqrt x + 3}} + 1 < 0\\
\Leftrightarrow \dfrac{{\left( {3\sqrt x – 9} \right) + \left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} < 0\\
\Leftrightarrow \dfrac{{4\sqrt x – 6}}{{\sqrt x + 3}} < 0\\
\Leftrightarrow 4\sqrt x – 6 < 0\\
\Leftrightarrow \sqrt x < \dfrac{3}{2}\\
\Leftrightarrow x < \dfrac{9}{4}\\
c,\\
P = \dfrac{{3\left( {\sqrt x – 3} \right)}}{{\sqrt x + 3}} = \dfrac{{3.\sqrt x – 9}}{{\sqrt x + 3}} = \dfrac{{3.\left( {\sqrt x + 3} \right) – 18}}{{\sqrt x + 3}} = 3 – \dfrac{{18}}{{\sqrt x + 3}}\\
P \in Z \Leftrightarrow \dfrac{{18}}{{\sqrt x + 3}} \in Z \Leftrightarrow \sqrt x + 3 \in \left\{ { \pm 1; \pm 2; \pm 3; \pm 6; \pm 9; \pm 18} \right\}\\
x > 4 \Rightarrow \sqrt x > 2 \Rightarrow \sqrt x + 3 > 5\\
\Rightarrow \sqrt x + 3 \in \left\{ {6;9;18} \right\}\\
\Rightarrow \sqrt x \in \left\{ {3;6;15} \right\}\\
\Rightarrow x \in \left\{ {9;36;\,225} \right\}\\
x \ne 9 \Rightarrow x \in \left\{ {36;225} \right\}
\end{array}\)