Đáp án:
$a)P=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\\ b)P(4-2\sqrt{3})=\dfrac{5-2\sqrt{3}}{13}\\ c)0 \le x <9, x \ne 1 \\ d)x \in \{4;16;25\}$
Giải thích các bước giải:
$a)P=\left[\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right]:\left[\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right]\\ =\left[\dfrac{x+2\sqrt{x}-7}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\right]:\left[\dfrac{\sqrt{x}-1}{(\sqrt{x}+3)(\sqrt{x}-1)}-\dfrac{\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}+3)}\right]\\ =\left[\dfrac{x+2\sqrt{x}-7}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{(\sqrt{x}-1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\dfrac{\sqrt{x}-1-(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}\\ =\dfrac{x+2\sqrt{x}-7-(\sqrt{x}-1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\dfrac{-4}{(\sqrt{x}-1)(\sqrt{x}+3)}\\ =\dfrac{-4}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{(\sqrt{x}-1)(\sqrt{x}+3)}{-4}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\\ b)x=4-2\sqrt{3}\\ =3-2\sqrt{3}+1\\ =(\sqrt{3}-1)^2\\ P(4-2\sqrt{3})\\ =\dfrac{\sqrt{(\sqrt{3}-1)^2}-1}{\sqrt{(\sqrt{3}-1)^2}-3}\\ =\dfrac{\sqrt{3}-1-1}{\sqrt{3}-1-3}\\ =\dfrac{\sqrt{3}-2}{\sqrt{3}-4}\\ =\dfrac{(\sqrt{3}-2)(\sqrt{3}+4)}{(\sqrt{3}-4)(\sqrt{3}+4)}\\ =\dfrac{5-2\sqrt{3}}{13}\\ c)P<1\\ \Leftrightarrow \dfrac{\sqrt{x}-1}{\sqrt{x}-3}<1\\ \Leftrightarrow \dfrac{\sqrt{x}-1}{\sqrt{x}-3}-1<0\\ \Leftrightarrow \dfrac{\sqrt{x}-1-(\sqrt{x}-3)}{\sqrt{x}-3}<0\\ \Leftrightarrow \dfrac{2}{\sqrt{x}-3}<0\\ \Leftrightarrow \sqrt{x}-3<0\\ \Leftrightarrow \sqrt{x}<3\\ \Leftrightarrow 0 \le x <9$
Kết hợp điều kiện $\Rightarrow 0 \le x <9, x \ne 1$
$d)P=\dfrac{\sqrt{x}-1}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow P=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow P =1+\dfrac{2}{\sqrt{x}-3} \in \mathbb{Z}\\ \Rightarrow \dfrac{2}{\sqrt{x}-3} \in \mathbb{Z}\\ \Rightarrow ( \sqrt{x}-3) \in Ư(2)\\ \Rightarrow x \in \{1;4;16;25\}$
Kết hợp điều kiện $\Rightarrow x \in \{4;16;25\}$
