Đáp án:

$1)P=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\\ 2)P(25)=\dfrac{11}{6}\\ 3)P(4-2\sqrt{3})=\dfrac{5\sqrt{3}+3}{3}\\ 4)P\left(\dfrac{2}{2+2\sqrt{3}}\right)=\dfrac{\sqrt{1-\sqrt{3}}+6\sqrt{2}}{\sqrt{1-\sqrt{3}}+\sqrt{2}}\\ 5) x=4\\ 6) x>\dfrac{4}{9}, x \ne 1\\ 7) \forall x \ge 0; x \ne 1 \\8)x \in \varnothing$

Giải thích các bước giải:

$P=\left(\dfrac{x+6}{\sqrt{x}-1}-\sqrt{x}\right)\left(2-\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\right)\\ ĐKXĐ: \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}-1 \ne 0  \\ \sqrt{x}+1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\  x \ne 1\end{array} \right.\\ 1)P=\left(\dfrac{x+6}{\sqrt{x}-1}-\sqrt{x}\right)\left(2-\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\right)\\ =\dfrac{x+6-\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}.\dfrac{2(\sqrt{x}+1)-(\sqrt{x}+3)}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+6}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\\ 2)P(25)=\dfrac{\sqrt{25}+6}{\sqrt{25}+1}=\dfrac{5+6}{5+1}=\dfrac{11}{6}\\ 3)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}+6}{\sqrt{(\sqrt{3}-1)^2}+1}\\ =\dfrac{\sqrt{3}-1+6}{\sqrt{3}-1+1}\\ =\dfrac{\sqrt{3}+5}{\sqrt{3}}\\ =\dfrac{5\sqrt{3}+3}{3}\\ 4)x=\dfrac{2}{2+2\sqrt{3}}\\ =\dfrac{1}{1+\sqrt{3}}\\ =\dfrac{(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\ =\dfrac{1-\sqrt{3}}{2}\\ P\left(\dfrac{2}{2+2\sqrt{3}}\right)\\ =\dfrac{\sqrt{\dfrac{1-\sqrt{3}}{2}}+6}{\sqrt{\dfrac{1-\sqrt{3}}{2}}+1}\\ =\dfrac{\sqrt{1-\sqrt{3}}+6\sqrt{2}}{\sqrt{1-\sqrt{3}}+\sqrt{2}}\\ 5)3P=2\sqrt{x}+4\\ \Leftrightarrow \dfrac{3(\sqrt{x}+6)}{\sqrt{x}+1}=2\sqrt{x}+4\\ \Leftrightarrow \dfrac{3(\sqrt{x}+6)}{\sqrt{x}+1}-(2\sqrt{x}+4)=0\\ \Leftrightarrow \dfrac{3\sqrt{x}+18}{\sqrt{x}+1}-\dfrac{(2\sqrt{x}+4)(\sqrt{x}+1)}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{3\sqrt{x}+18-(2\sqrt{x}+4)(\sqrt{x}+1)}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{-2x-3\sqrt{x}+14}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{2x+3\sqrt{x}-14}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{2x-4\sqrt{x}+7\sqrt{x}-14}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{2\sqrt{x}(\sqrt{x}-2)+7(\sqrt{x}-2)}{\sqrt{x}+1}=0\\ \Leftrightarrow \dfrac{(2\sqrt{x}+7)(\sqrt{x}-2)}{\sqrt{x}+1}=0\\ \Leftrightarrow \sqrt{x}-2=0(Do \ \ 2\sqrt{x}+7 >0 ; \sqrt{x}+1 >0 \ \forall x \ge 0; x \ne 1)\\ \Leftrightarrow x=4\\ 6)P<4\\ \Leftrightarrow \dfrac{\sqrt{x}+6}{\sqrt{x}+1}<4\\ \Leftrightarrow \dfrac{\sqrt{x}+6}{\sqrt{x}+1}-4<0\\ \Leftrightarrow \dfrac{\sqrt{x}+6-4(\sqrt{x}+1)}{\sqrt{x}+1}<0\\ \Leftrightarrow \dfrac{2-3\sqrt{x}}{\sqrt{x}+1}<0\\ \Leftrightarrow 2-3\sqrt{x}<0 (Do \ \sqrt{x}+1 >0 \ \forall x \ge 0; x \ne 1)\\ \Leftrightarrow \sqrt{x}>\dfrac{2}{3}\\ \Leftrightarrow x>\dfrac{4}{9}\\ \text{Kết hợp điều kiện} \Rightarrow x>\dfrac{4}{9}, x \ne 1\\ 7)P>-\dfrac{5}{3}\\ P=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\\ Do \ \ \sqrt{x}+6 >0 ; \sqrt{x}+1>0 \ \forall x \ge 0; x \ne 1)\\ \Rightarrow P>0 \ \forall x \ge 0; x \ne 1\\ \Rightarrow P >-\dfrac{5}{3} \ \forall x \ge 0; x \ne 1\\ 8)P \le \dfrac{2}{5}\\ P= \dfrac{\sqrt{x}+6}{\sqrt{x}+1}\\ = \dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}\\ =1+\dfrac{5}{\sqrt{x}+1} > \dfrac{2}{5}\ \forall x \ge 0; x \ne 1\\ \Rightarrow P > \dfrac{2}{5} \ \forall x \ge 0; x \ne 1$

$\Rightarrow$ Không tồn tại $x$ để $P \le \dfrac{2}{5}$