a, đk x >= 0 ; x khác 1
\(P=\dfrac{15\sqrt{x}-11+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)+\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}-3}\)
\(=\dfrac{15\sqrt{x}-11+3x+7\sqrt{x}-6+2x+\sqrt{x}-3}{x+2\sqrt{x}-3}=\dfrac{23\sqrt{x}+5x-20}{x+2\sqrt{x}-3}\)
b, Thay x = 9 ta được
\(\dfrac{23.3+5.9-20}{9+2.3-3}=\dfrac{94}{12}=\dfrac{47}{6}\)
c, \(5x+23\sqrt{x}-20=2x+4\sqrt{x}-6\Leftrightarrow3x+19\sqrt{x}-14=0\Leftrightarrow\sqrt{x}=\dfrac{2}{3};\sqrt{x}=-7\left(voli\right)\Rightarrow x=\dfrac{4}{9}\left(tm\right)\)
a. \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2-3\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11+\left(2-3\sqrt{x}\right)\left(3+\sqrt{x}\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11+6+2\sqrt{x}-9\sqrt{x}-3x-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{5x-7\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b. Thay x = 9 (tmđk) vào P :
\(P=\dfrac{2-5\sqrt{9}}{\sqrt{9}+3}=-\dfrac{13\sqrt{3}}{6}\)
Vậy : Với x = 9 thì \(P=-\dfrac{13\sqrt{3}}{6}\)
c. \(P=\dfrac{1}{2}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow4-10\sqrt{x}=\sqrt{x}+3\)
\(\Leftrightarrow11\sqrt{x}=1\)
\(\Leftrightarrow x=\dfrac{1}{121}\left(tmđk\right)\)
Vậy : Với \(x=\dfrac{1}{121}\) thì \(P=\dfrac{1}{2}\)
d. Ta có : \(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{5\left(\sqrt{x}+3\right)-13}{\sqrt{x}+3}=5-\dfrac{13}{\sqrt{x}+3}\)
\(P\in Z\Rightarrow\left\{{}\begin{matrix}5\in Z\\\dfrac{13}{\sqrt{x}+3}\in Z\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Ta có bảng :
| \(\sqrt{x}+3\) | 1 | -1 | 13 | -13 |
| \(x\) | || | || | 100 | || |
| TMĐK | --- | --- | Nhận | --- |
Vậy : P nguyên khi x = 100.
