Đáp án:
$3)A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b)A=\dfrac{3-\sqrt{5}}{4}\\ c) A<2\\ d)max_P= -5 \Leftrightarrow x=\dfrac{1}{9}$
Giải thích các bước giải:
$3)A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)^2}\\ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ x-\sqrt{x} \ne 0\\ \sqrt{x}-1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1\end{array} \right.\\ A=\left(\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}+1}\\ =\left(\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right).\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}.\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b)A(6-2\sqrt{5})\\ =\dfrac{\sqrt{6-2\sqrt{5}}-1}{\sqrt{6-2\sqrt{5}}}\\ =\dfrac{\sqrt{5-2\sqrt{5}+1}-1}{\sqrt{5-2\sqrt{5}+1}}\\ =\dfrac{\sqrt{(\sqrt{5}-1)^2}-1}{\sqrt{(\sqrt{5}-1)^2}}\\ =\dfrac{|\sqrt{5}-1|-1}{|\sqrt{5}-1|}\\ =\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1}\\ =\dfrac{\sqrt{5}-2}{\sqrt{5}-1}\\ =\dfrac{(\sqrt{5}-2)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\\ =\dfrac{3-\sqrt{5}}{5-1}\\ =\dfrac{3-\sqrt{5}}{4}\\ c)A-2\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}-2\\ =\dfrac{\sqrt{x}-1-2\sqrt{x}}{\sqrt{x}}\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}} <0 \ \forall \ x> 0; x \ne 1\\ \Rightarrow A<2\\ d)P=A-9\sqrt{x}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}\\ =\dfrac{\sqrt{x}-1-9x}{\sqrt{x}}\\ =\dfrac{\sqrt{x}-1-9x}{\sqrt{x}}\\ =\dfrac{-9x+6\sqrt{x}-1-5\sqrt{x}}{\sqrt{x}}\\ =-\dfrac{(3\sqrt{x}-1)^2}{\sqrt{x}}-5 \le -5 \ \forall \ x> 0; x \ne 1$
Dấu "=" xảy ra $\Leftrightarrow 3\sqrt{x}-1=0 \Leftrightarrow x=\dfrac{1}{9}$
