Đáp án:

$4)\\ a)A=4a\\ b)A(\sqrt{2})=4\sqrt{2}\\ c)a>1\\ d)min_M=16 \Leftrightarrow a=4$

Giải thích các bước giải:

$4)\\ A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\\ ĐKXĐ: \left\{\begin{array}{l} a \ge 0 \\ \sqrt{a}-1 \ne 0 \\ \sqrt{a}+1 \ne 0 \\ \sqrt{a} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a > 0 \\ a \ne 1 \end{array} \right.\\ a)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{(\sqrt{a}+1)^2}{(\sqrt{a}-1)(\sqrt{a}+1)}-\dfrac{(\sqrt{a}-1)^2}{(\sqrt{a}-1)(\sqrt{a}+1)}+\dfrac{4\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)(\sqrt{a}+1)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\dfrac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+4\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)(\sqrt{a}+1)}.\dfrac{a-1}{\sqrt{a}}\\ =\dfrac{4a\sqrt{a}}{a-1}.\dfrac{a-1}{\sqrt{a}}\\ =4a\\ b)a=(2+\sqrt{3})(\sqrt{3}-1)\sqrt{2-\sqrt{3}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)(\sqrt{3}-1)}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(\sqrt{3}-1)^2}{\sqrt{2}}\\ =\dfrac{(2+\sqrt{3})(4-2\sqrt{3})}{\sqrt{2}}\\ =\dfrac{2(2+\sqrt{3})(2-\sqrt{3})}{\sqrt{2}}\\ =\sqrt{2}(4-3)\\ =\sqrt{2}\\ A(\sqrt{2})=4\sqrt{2}\\ c)\dfrac{A}{4(\sqrt{a}-1)}>1\\ \Leftrightarrow \dfrac{4a}{4(\sqrt{a}-1)}-1>0\\ \Leftrightarrow \dfrac{a}{\sqrt{a}-1}-1>0\\ \Leftrightarrow \dfrac{a-(\sqrt{a}-1)}{\sqrt{a}-1}>0\\ \Leftrightarrow \dfrac{a-\sqrt{a}+1}{\sqrt{a}-1}>0\\ \Leftrightarrow \dfrac{a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}{\sqrt{a}-1}>0\\ \Leftrightarrow \dfrac{\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\sqrt{a}-1}>0\\ \Rightarrow \sqrt{a}-1>0 \left(Do \ \left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4} >0 \ \forall \ x \right)\\ \Leftrightarrow \sqrt{a}>1\\ \Leftrightarrow a>1\\ d)M=\dfrac{A}{\sqrt{a}-1}\\ =\dfrac{4a}{\sqrt{a}-1}\\ =\dfrac{4a-4\sqrt{a}+4\sqrt{a}-4+4}{\sqrt{a}-1}\\ =\dfrac{4\sqrt{a}(\sqrt{a}-1)+4(\sqrt{a}-1)+4}{\sqrt{a}-1}\\ =4\sqrt{a}+4+\dfrac{4}{\sqrt{a}-1}\\ =4\sqrt{a}-4+\dfrac{4}{\sqrt{a}-1}+8\\ =4(\sqrt{a}-1)+\dfrac{4}{\sqrt{a}-1}+8(\sqrt{a}-1>0 \ do \ a>1)\\ \ge 2\sqrt{4(\sqrt{a}-1).\dfrac{4}{\sqrt{a}-1}}+8 (Cauchy)\\ = 16$

Dấu "=" xảy ra

$\Leftrightarrow 4(\sqrt{a}-1)= \dfrac{4}{\sqrt{a}-1}\Leftrightarrow (\sqrt{a}-1)^2=1\Leftrightarrow a=0(L); a=4$