Đáp án:
$1)min_A=2\sqrt{13}\Leftrightarrow x=13\\ 2)min_B=-\dfrac{9}{13} \Leftrightarrow x=0\\ 3)\\ a)\text{ĐKXĐ:} \left\{\begin{array}{l} x > 0\\ x \ne 1 \end{array} \right.\\ b)M=\dfrac{2}{x-1}\\ c) 0<x<1$
Giải thích các bước giải:
$1)\\ A=\dfrac{x+13}{\sqrt{x}}(x>0)\\ =\sqrt{x}+\dfrac{13}{\sqrt{x}} \ge 2\sqrt{\sqrt{x}.\dfrac{13}{\sqrt{x}} }=2\sqrt{13}(Cauchy)$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=\dfrac{13}{\sqrt{x}} \Leftrightarrow x=13$
$2)\\ B=\dfrac{2\sqrt{x}-9}{\sqrt{x}+13}(x \ge 0)\\ =\dfrac{2\sqrt{x}+26-35}{\sqrt{x}+13}\\ =\dfrac{2(\sqrt{x}+13)-35}{\sqrt{x}+13}\\ =2-\dfrac{35}{\sqrt{x}+13}\\ \sqrt{x}+13 \ge 13 \ \forall \ x \ge 0 \\ \Rightarrow \dfrac{35}{\sqrt{x}+13} \le \dfrac{35}{13} \ \forall \ x \ge 0 \\ \Rightarrow -\dfrac{35}{\sqrt{x}+13} \ge -\dfrac{35}{13} \ \forall \ x \ge 0 \\ \Rightarrow 2-\dfrac{35}{\sqrt{x}+13} \ge 2 -\dfrac{35}{13}=-\dfrac{9}{13} \ \forall \ x \ge 0 $ Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=0 \Leftrightarrow x=0$ $3)\\ M=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ a)\text{ĐKXĐ:} \left\{\begin{array}{l} x \ge 0\\x+2\sqrt{x}+1 \ne 0 \\ x-1 \ne 0 \\ \sqrt{x} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\(\sqrt{x}+1)^2 \ne 0 \\ x \ne 1 \\ x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \end{array} \right.\\ b)M=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\dfrac{\sqrt{x}-2}{(\sqrt{x}+1)(\sqrt{x}-1)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ \\ =\dfrac{(\sqrt{x}+2)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{x-\sqrt{x}+2\sqrt{x}-2-(x+\sqrt{x}-2\sqrt{x}-2)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\=\dfrac{2\sqrt{x}}{(\sqrt{x}+1)^2(\sqrt{x}-1)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{2}{x-1}\\ c)|M+1|>M+1\\ \Leftrightarrow \left[\begin{array}{l} M+1<0\\ \left\{\begin{array}{l} M+1\ge 0\\ (M+1)^2 > (M+1)^2\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \dfrac{2}{x-1}+1<0\\ \left\{\begin{array}{l} M+1\ge 0\\ 0 > 0(\text{Vô lí})\end{array} \right.\end{array} \right.\\ \Leftrightarrow \dfrac{2}{x-1}+1<0 \\ \Leftrightarrow \dfrac{2+x-1}{x-1}<0 \\ \Leftrightarrow \dfrac{x+1}{x-1}<0 \\ \Leftrightarrow -1<x<1\\ \text{Kết hợp điều kiện} \Rightarrow 0<x<1$
