a/ ĐKXĐ: \(x>4\)
\(x=9(TM)→A=\dfrac{9+2011}{\sqrt 9-2}=2020\)
b/ \(\dfrac{\sqrt x-1}{\sqrt x+2}+\dfrac{5\sqrt x-2}{x-4}(x>4)\\=\dfrac{(\sqrt x-1)(\sqrt x-2)}{(\sqrt x-2)(\sqrt x+2)}+\dfrac{5\sqrt x-2}{(\sqrt x-2)(\sqrt x+2)}\\=\dfrac{x-3\sqrt x+2+5\sqrt x-2}{(\sqrt x+2)(\sqrt x-2)}\\=\dfrac{x+2\sqrt x}{(\sqrt x+2)(\sqrt x-2)}\\=\dfrac{\sqrt x(\sqrt x+2)}{(\sqrt x+2)(\sqrt x-2)}\\=\dfrac{\sqrt x}{\sqrt x-2}\)
c/ \(\dfrac{A}{B}=\dfrac{\dfrac{x+2011}{\sqrt x-2}}{\dfrac{\sqrt x}{\sqrt x-2}}=\dfrac{x+2011}{\sqrt x}\\=\sqrt x+\dfrac{2011}{\sqrt x}\)
Áp dụng BĐT Cô-si với 2 số dương
\(\sqrt x+\dfrac{2011}{\sqrt x}\ge 2\sqrt{\sqrt x.\dfrac{2011}{\sqrt x}}=2\sqrt{2011}\)
\(→\min =2\sqrt{2011}\)
\(→\) Dấu "=" xảy ra khi \(\sqrt x=\dfrac{2011}{\sqrt x}\\↔x=2011(TM)\)
Vậy \(\min =2\sqrt{2011}\)
