Đáp án:
$21)\\ a)\text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0\\x \ne 9 \\x \ne 4 \end{array} \right.\\ b)A=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c) x \in \{0;16;36\}\\ 22)\\ B=\sqrt{2a-1}+\sqrt{5}\\ 23)\\ a) x \in \{0;4;9\}\\ b)0 \le x <1\\ c) x \in \varnothing\\ d)$
Giải thích các bước giải:
$21)\\ a)A=\dfrac{x-4\sqrt{x}+1}{x-5\sqrt{x}+6}+\dfrac{5}{\sqrt{x}-3}-\dfrac{3}{\sqrt{x}-2}\\ \text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0\\ x-5\sqrt{x}+6 \ne 0\\ \sqrt{x}-3 \ne 0 \\\sqrt{x}-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x-2\sqrt{x}-3\sqrt{x}+6 \ne 0\\ \sqrt{x}-3 \ne 0 \\\sqrt{x}-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2) \ne 0\\ \sqrt{x} \ne 3 \\\sqrt{x} \ne 2 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ (\sqrt{x}-2)(\sqrt{x}-3) \ne 0\\ \sqrt{x} \ne 3 \\\sqrt{x} \ne 2 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ (\sqrt{x}-2)(\sqrt{x}-3) \ne 0\\ x \ne 9 \\x \ne 4 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\x \ne 9 \\x \ne 4 \end{array} \right.\\ b)A=\dfrac{x-4\sqrt{x}+1}{x-5\sqrt{x}+6}+\dfrac{5}{\sqrt{x}-3}-\dfrac{3}{\sqrt{x}-2} =\dfrac{x-4\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}-3)}+\dfrac{5}{\sqrt{x}-3}-\dfrac{3}{\sqrt{x}-2}\\ =\dfrac{x-4\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}-3)}+\dfrac{5(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}-\dfrac{3(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}\\ =\dfrac{x-4\sqrt{x}+1+5(\sqrt{x}-2)-3(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}\\ =\dfrac{x-2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c)A=\dfrac{3}{2}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{3}{2}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3}{2}=0\\ \Leftrightarrow \dfrac{2\sqrt{x}-3(\sqrt{x}-3)}{2(\sqrt{x}-3)}=0\\ \Leftrightarrow \dfrac{-\sqrt{x}+9}{2(\sqrt{x}-3)}=0\\ \Leftrightarrow -\sqrt{x}+9=0\\ \Leftrightarrow \sqrt{x}=9\\ \Leftrightarrow x=81\\ d)A=\dfrac{\sqrt{x}}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}-3+3}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow 1+\dfrac{3}{\sqrt{x}-3} \in \mathbb{Z}\\ \Rightarrow \dfrac{3}{\sqrt{x}-3} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}-3) \in Ư(3)\\ \Leftrightarrow (\sqrt{x}-3) \in \{\pm 1;\pm 3\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-3=-3\\ \sqrt{x}-3=-1 \\ \sqrt{x}-3=1\\ \sqrt{x}-3=3\end{array} \right. \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=0\\ \sqrt{x}=2 \\ \sqrt{x}=4\\ \sqrt{x}=6\end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=0\\ x=4(L) \\ x=16\\ x=36\end{array} \right. \Rightarrow x \in \{0;16;36\}\\ 22)\\ B=\dfrac{2a-6}{\sqrt{2a-1}-\sqrt{5}}\\ =\dfrac{(2a-6)\left(\sqrt{2a-1}+\sqrt{5}\right)}{\left(\sqrt{2a-1}-\sqrt{5}\right)\left(\sqrt{2a-1}+\sqrt{5}\right)}\\ =\dfrac{(2a-6)\left(\sqrt{2a-1}+\sqrt{5}\right)}{2a-1-5}\\ =\dfrac{(2a-6)\left(\sqrt{2a-1}+\sqrt{5}\right)}{2a-6}\\ =\sqrt{2a-1}+\sqrt{5}\\ 23)\\ Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ \text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0\\ \sqrt{x}-1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x} \ne 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x \ne 1\end{array} \right.\\ a)Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ \Leftrightarrow Q=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}\\ \Leftrightarrow Q=1+\dfrac{2}{\sqrt{x}-1}\\ \Rightarrow \dfrac{2}{\sqrt{x}-1} \in \mathbb{Z} \\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}-1) \in Ư(2)\\ \Leftrightarrow (\sqrt{x}-1) \in \{\pm1;\pm2\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-1=-2\\ \sqrt{x}-1=-1 \\ \sqrt{x}-1=1\\ \sqrt{x}-1=2\end{array} \right. \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=-1\\ \sqrt{x}=0 \\ \sqrt{x}=2\\ \sqrt{x}=3\end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=0 \\ x=4\\ x=9\end{array} \right.\Rightarrow x \in \{0;4;9\}\\ b)Q<0\\ \Leftrightarrow \dfrac{\sqrt{x}+1}{\sqrt{x}-1}<0\\ \Leftrightarrow \sqrt{x}-1<0(\text{Do }\sqrt{x}+1>0 \ \forall x\ge 0, x \ne 1)\\ \Leftrightarrow \sqrt{x}<1\\ \Leftrightarrow0 \le x <1\\ c)Q=1\\ \Leftrightarrow \dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1\\ \Leftrightarrow \dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1=0\\ \Leftrightarrow \dfrac{\sqrt{x}+1-(\sqrt{x}-1)}{\sqrt{x}-1}=0\\ \Leftrightarrow \dfrac{2}{\sqrt{x}-1}=0\\ \Leftrightarrow x \in \varnothing\\ d)$
