Đáp án:
$a)A=\dfrac{2\sqrt{x}-4}{x}\\ b)A=3\sqrt{3}-5\\ c)x>0,x\ne 16\\ d)x=16$
Giải thích các bước giải:
$a)ĐKXĐ:x \ne 0, x\ne 4\\ A=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}\right):\dfrac{x\sqrt{x}}{(4-x)^2}\\ =\left(\dfrac{\sqrt{x}-1}{(\sqrt{x}-2)(\sqrt{x}+2)}-\dfrac{\sqrt{x}+1}{(\sqrt{x}+2)^2}\right).\dfrac{(4-x)^2}{x\sqrt{x}}\\ =\dfrac{(\sqrt{x}-1)(\sqrt{x}+2)-(\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)^2}.\dfrac{(\sqrt{x}-2)^2(\sqrt{x}+2)^2}{x\sqrt{x}}\\ =\dfrac{2\sqrt{x}-4}{x}\\ b)x=4+2\sqrt{3}=3+2\sqrt{3}+1=(\sqrt{3}+1)^2\\ A(4+2\sqrt{3})\\ =\dfrac{2\sqrt{4+2\sqrt{3}}-4}{4+2\sqrt{3}}\\ =\dfrac{2\sqrt{3+2\sqrt{3}+1}-4}{4+2\sqrt{3}}\\ =\dfrac{2\sqrt{(\sqrt{3}+1)^2}-4}{4+2\sqrt{3}}\\ =\dfrac{2(\sqrt{3}+1)-4}{4+2\sqrt{3}}\\ =\dfrac{2\sqrt{3}-2}{4+2\sqrt{3}}\\ =\dfrac{\sqrt{3}-1}{2+\sqrt{3}}\\ =\dfrac{(\sqrt{3}-1)(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}\\ =\dfrac{3\sqrt{3}-5}{4-3}\\ =3\sqrt{3}-5\\ c)A<\dfrac{1}{4}\\ \Leftrightarrow \dfrac{2\sqrt{x}-4}{x}<\dfrac{1}{4}\\ \Leftrightarrow \dfrac{2\sqrt{x}-4}{x}-\dfrac{1}{4}<0\\ \Leftrightarrow \dfrac{8\sqrt{x}-16-x}{4x}<0\\ \Leftrightarrow \dfrac{-(\sqrt{x}-4)^2}{4x}<0\\ \circledast -(\sqrt{x}-4)^2=0 \Leftrightarrow x=16 \Rightarrow A=0(L)\\ \circledast -(\sqrt{x}-4)^2\ne 0\\ \Leftrightarrow \dfrac{-(\sqrt{x}-4)^2}{4x}<0\\ \Leftrightarrow 4x>0\\ \Leftrightarrow x>0$
Vậy với $x>0, x \ne 16$ thì $A<\dfrac{1}{4}$
$d)A=\dfrac{2\sqrt{x}-4}{x}\\ =\dfrac{-4}{x}+\dfrac{2}{\sqrt{x}}\\ =-\left(\dfrac{4}{x}-\dfrac{2}{\sqrt{x}}\right)\\ =-\left(\left(\dfrac{2}{\sqrt{x}}\right)^2-2.\dfrac{2}{\sqrt{x}}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{4}\\ =-\left(\dfrac{2}{\sqrt{x}}-\dfrac{1}{2}\right)^2+\dfrac{1}{4} \ge \dfrac{1}{4}$
Dấu "=" xảy ra $\Leftrightarrow \dfrac{2}{\sqrt{x}}-\dfrac{1}{2}=0$
$\Leftrightarrow \dfrac{2}{\sqrt{x}}=\dfrac{1}{2}\\ \Leftrightarrow x=16$
