giúp mình với mình đang cần gấp
cho 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 $\sqrt{x}$ = $\frac{\sqrt{3}+1}{\sqrt{2}}$
Đáp án:
b. \(x > \dfrac{9}{4};x \ne 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
P = \left[ {\dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) – 3\sqrt x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x – 2 – \sqrt x + 3}}{{\sqrt x – 3}}} \right)\\
= \left[ {\dfrac{{2x – 6\sqrt x + x + 3\sqrt x – 3\sqrt x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right].\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{{3x – 6\sqrt x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\
= \dfrac{{3x – 6\sqrt x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{3x – 6\sqrt x – 9}}{{x + 4\sqrt x + 3}}\\
b.P < – 1\\
\to \dfrac{{3x – 6\sqrt x – 9}}{{x + 4\sqrt x + 3}} < – 1\\
\to \dfrac{{3x – 6\sqrt x – 9 + x + 4\sqrt x + 3}}{{x + 4\sqrt x + 3}} < 0\\
\to \dfrac{{4x – 2\sqrt x – 6}}{{x + 4\sqrt x + 3}} < 0\\
\to \dfrac{{ – \left( {2\sqrt x – 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{2\sqrt x – 3}}{{\sqrt x + 3}} > 0\\
\to 2\sqrt x – 3 > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0;x \ne 9} \right)\\
\to x > \dfrac{9}{4};x \ne 9
\end{array}\)