Đáp án:

$a)P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\\ b) x=4\\ c) x=0\\ d)P(13-4\sqrt{10})=\dfrac{273 +4 \sqrt 2 - 14 \sqrt 5 + 16 \sqrt 10}{79}\\ e) 0 \le x \le 16, x \ne 1$

Giải thích các bước giải:

$P=\dfrac{3(x+\sqrt{x}-3)}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\\ \text{ĐKXĐ:}  \left\{\begin{array}{l} x \ge 0 \\ x+\sqrt{x}-2 \ne 0 \\ \sqrt{x}+2 \ne 0 \\\sqrt{x}-1 \ne 0\end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ge 0 \\ x+2\sqrt{x}-\sqrt{x}-2 \ne 0 \\ \sqrt{x} \ne -2 \\\sqrt{x} \ne 1\end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}(\sqrt{x}+2)-(\sqrt{x}+2) \ne 0 \\ x\ne 1\end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ge 0 \\ (\sqrt{x}-1)(\sqrt{x}+2)\ne 0 \\ x\ne 1\end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ge 0 \\ x\ne 1\end{array} \right.\\ a)\dfrac{3(x+\sqrt{x}-3)}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\\ =\dfrac{3(x+\sqrt{x}-3)}{ (\sqrt{x}-1)(\sqrt{x}+2)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\\ =\dfrac{3(x+\sqrt{x}-3)}{ (\sqrt{x}-1)(\sqrt{x}+2)}+\dfrac{(\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{3(x+\sqrt{x}-3)+(\sqrt{x}+3)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{3x+5\sqrt{x}-8}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{3x-3\sqrt{x}+8\sqrt{x}-8}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{3\sqrt{x}(\sqrt{x}-1)+8(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{(\sqrt{x}-1)(3\sqrt{x}+8)}{(\sqrt{x}-1)(\sqrt{x}+2)}\\ =\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\\ b)P=\dfrac{7}{2}\\ \Leftrightarrow \dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{7}{2}\\ \Leftrightarrow \dfrac{3\sqrt{x}+8}{\sqrt{x}+2}-\dfrac{7}{2}=0\\ \Leftrightarrow \dfrac{2(3\sqrt{x}+8)-7(\sqrt{x}+2)}{2(\sqrt{x}+2)}=0\\ \Leftrightarrow \dfrac{2-\sqrt{x}}{2(\sqrt{x}+2)}=0\\ \Leftrightarrow 2-\sqrt{x}=0\\ \Leftrightarrow \sqrt{x}=2\\ \Leftrightarrow x=4\\ c)P \in \mathbb{Z}\\ \Leftrightarrow \dfrac{3\sqrt{x}+8}{\sqrt{x}+2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{3(\sqrt{x}+2)+2}{\sqrt{x}+2} \in \mathbb{Z}\\ \Leftrightarrow 3+\dfrac{2}{\sqrt{x}+2} \in \mathbb{Z}\\ \Rightarrow \dfrac{2}{\sqrt{x}+2} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}+2) \in Ư(2)\\ \Leftrightarrow (\sqrt{x}+2) \in \{\pm1;\pm2\}\\ \Leftrightarrow  \left\{\begin{array}{l} \sqrt{x}+2=-2\\ \sqrt{x}+2=-1 \\ \sqrt{x}+2=1\\\sqrt{x}+2=2\end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} \sqrt{x}=-4\\ \sqrt{x}=-3\\ \sqrt{x}=-1\\\sqrt{x}=0\end{array} \right.\\ \Leftrightarrow x=0\\ d)x=13-4\sqrt{10}\\ =8-2.2\sqrt{2}.\sqrt{5}+5\\ =(2\sqrt{2}-\sqrt{5})^2\\ P(13-4\sqrt{10})=\dfrac{3\sqrt{(2\sqrt{2}-\sqrt{5})^2}+8}{\sqrt{(2\sqrt{2}-\sqrt{5})^2}+2}\\ =\dfrac{3(2\sqrt{2}-\sqrt{5})+8}{2\sqrt{2}-\sqrt{5}+2}\\ =\dfrac{6\sqrt{2}-3\sqrt{5}+8}{2\sqrt{2}-\sqrt{5}+2}\\ =\dfrac{(6\sqrt{2}-3\sqrt{5}+8)(2\sqrt{2}+2+\sqrt{5})}{(2\sqrt{2}+2-\sqrt{5})(2\sqrt{2}+2+\sqrt{5})}\\ =\dfrac{24+12\sqrt{2}+6\sqrt{10}-6\sqrt{10}-6\sqrt{5}-15+16\sqrt{2}+16+8\sqrt{5}}{(2\sqrt{2}+2)^2-5}\\ =\dfrac{25+28\sqrt{2}+2\sqrt{5}}{8+8\sqrt{2}+4-5}\\ =\dfrac{25+28\sqrt{2}+2\sqrt{5}}{7+8\sqrt{2}}\\ =\dfrac{(25+28\sqrt{2}+2\sqrt{5})(7-8\sqrt{2})}{(7+8\sqrt{2})(7-8\sqrt{2})}\\ =\dfrac{175+196\sqrt{2}+14\sqrt{5}-200\sqrt{2}-448-16\sqrt{10}}{(7+8\sqrt{2})(7-8\sqrt{2})}\\ =\dfrac{-273 - 4 \sqrt 2 + 14 \sqrt 5 - 16 \sqrt 10}{49-128}\\ =\dfrac{273 +4 \sqrt 2 - 14 \sqrt 5 + 16 \sqrt 10}{79}\\ e)P>\dfrac{10}{3}\\ \Leftrightarrow \dfrac{3\sqrt{x}+8}{\sqrt{x}+2}>\dfrac{10}{3}\\ \Leftrightarrow \dfrac{3\sqrt{x}+8}{\sqrt{x}+2}-\dfrac{10}{3}>0\\ \Leftrightarrow \dfrac{3(3\sqrt{x}+8)-10(\sqrt{x}+2)}{3(\sqrt{x}+2)}>0\\ \Leftrightarrow \dfrac{4-\sqrt{x}}{3(\sqrt{x}+2)}>0\\ \Leftrightarrow 4-\sqrt{x}>0\\ \Leftrightarrow \sqrt{x}<4\\ \Leftrightarrow 0 \le x \le 16\\ \text{Kết hợp điều kiện} \Rightarrow 0 \le x \le 16, x \ne 1$