Lời giải:
a) ĐKXĐ: \(a>0; a\neq 1\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+4\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)(\sqrt{a}+1)}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{a+1+2\sqrt{a}-(a+1-2\sqrt{a})+4\sqrt{a}(a-1)}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4\sqrt{a}+4\sqrt{a}(a-1)}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4\sqrt{a}.a}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a(a+1)}{a-1}\)
b)
Ta có:
\(a=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(16-15)=2\)
Thay $a=2$ vào biểu thức đã thu gọn:
\(A=24\)
