Đáp án:

$8)\\  a) P=\left(a-1\right)^2\\ b) \left\{\begin{array}{l}-1+\sqrt{3}<a<3-\sqrt{3} \\ a \ne 1\end{array} \right.\\ 9) a)P=\dfrac{-3}{\sqrt{x}+3}\\ b) \forall \ x \ge 0 ,x \ne 9\\ c)min_P=-1 \Leftrightarrow 0$

Giải thích các bước giải:

$8)\\ P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\\ \text{ĐKXĐ}: \left\{\begin{array}{l}a \ge 0 \\1-\sqrt{a} \ne 0 \\ 1+\sqrt{a} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}a \ge 0 \\\sqrt{a} \ne 1 \\ \sqrt{a} \ne -1 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}a \ge 0 \\ a \ne 1\end{array} \right.\\ a) P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\dfrac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\dfrac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(1-\sqrt{a}+a-\sqrt{a}\right)\\ =\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\\ =\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\\ =\left(\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\right)^2\\ =\left(a-1\right)^2\\ b)P<7-4\sqrt{3}\\ \Leftrightarrow \left(a-1\right)^2 <7-4\sqrt{3}\\ \Leftrightarrow \left(a-1\right)^2 <4-2.2\sqrt{3}+3\\ \Leftrightarrow \left(a-1\right)^2 <(2-\sqrt{3})^2\\ \Leftrightarrow \left(a-1\right)^2- (2-\sqrt{3})^2<0\\ \Leftrightarrow \left(a-1-2+\sqrt{3}\right)\left(a-1+2-\sqrt{3}\right)<0\\ \Leftrightarrow \left(a-3+\sqrt{3}\right)\left(a+1-\sqrt{3}\right)<0\\ \Leftrightarrow \left(a-(3-\sqrt{3})\right)\left(a-(-1+\sqrt{3})\right)<0\\ \Leftrightarrow -1+\sqrt{3}<a<3-\sqrt{3}\\ \text{Kết hợp điều kiện}\Rightarrow \left\{\begin{array}{l}-1+\sqrt{3}<a<3-\sqrt{3} \\ a \ge 0 \\ a \ne 1\end{array} \right. \Leftrightarrow \left\{\begin{array}{l}-1+\sqrt{3}<a<3-\sqrt{3} \\ a \ne 1\end{array} \right.\\ 9)\\ P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\\ \text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}+3 \ne 0\\\sqrt{x}-3 \ne 0 \\x- 9\ne 0 \\ \left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x} \ne -3\\\sqrt{x} \ne 3 \\x\ne9 \\ \dfrac{2\sqrt{x}-2-(\sqrt{x}-3)}{\sqrt{x}-3} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 9 \\ \dfrac{\sqrt{x}+1}{\sqrt{x}-3} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 9 \\ \sqrt{x}+1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 9 \end{array} \right.\\ a)P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{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{3x+3}{(\sqrt{x}-3)(\sqrt{x}+3)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\left(\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{3x+3}{(\sqrt{x}-3)(\sqrt{x}+3)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-(3x+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3(\sqrt{x}+1)}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\\ b)P<\dfrac{1}{2}\\ \Leftrightarrow \dfrac{-3}{\sqrt{x}+3}<\dfrac{1}{2}\\ \Leftrightarrow \dfrac{-3}{\sqrt{x}+3}-\dfrac{1}{2}<0\\ \Leftrightarrow \dfrac{-3.2-(\sqrt{x}+3)}{2(\sqrt{x}+3)}<0\\ \Leftrightarrow \dfrac{-9-\sqrt{x}}{2(\sqrt{x}+3)}<0$

Do $-9-\sqrt{x}<0 \ \forall \ x \ge 0 , x \ne 9;2(\sqrt{x}+3)>0 \ \forall \ x \ge 0 ,x \ne 9$

$\Rightarrow \dfrac{-9-\sqrt{x}}{2(\sqrt{x}+3)}<0 \ \forall \ x \ge 0 ,x \ne 9\\ \Rightarrow P<\dfrac{1}{2} \ \forall \ x \ge 0 ,x \ne 9\\ c)P=\dfrac{-3}{\sqrt{x}+3}\\ \sqrt{x}+3 \ge 3 \ \forall \ x \ge 0 ,x \ne 9\\ \Rightarrow \dfrac{-3}{\sqrt{x}+3} \ge \dfrac{-3}{3}=-1 \ \forall \ x \ge 0 ,x \ne 9$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=0  \Leftrightarrow 0$