a/ ĐKXĐ: \(x>0;x\ne 1\)
\(A=(\dfrac{1}{\sqrt x+1}+\dfrac{2}{x-1}).(\dfrac{\sqrt x}{\sqrt x+1}-\dfrac{1}{x+\sqrt x})\\=(\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{2}{(\sqrt x-1)(\sqrt x+1)}).(\dfrac{x}{\sqrt x(\sqrt x+1)}-\dfrac{1}{\sqrt x(\sqrt x+1)})\\=\dfrac{\sqrt x-1+2}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{x-1}{\sqrt x(\sqrt x+1)}\\=\dfrac{\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}.\dfrac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x(\sqrt x+1)}\\=\dfrac{1}{\sqrt x-1}.\dfrac{\sqrt x-1}{\sqrt x}\\=\dfrac{1}{\sqrt x}\)
b/ \(x=4-2\sqrt 3\\=3-2\sqrt 3+1\\=(\sqrt 3-1)^2\)
Thay \(x=(\sqrt 3-1)^2(TM)\) vào biểu thức \(A\)
\(→A=\dfrac{1}{\sqrt{(\sqrt 3-1)^2}}\\=\dfrac{1}{|\sqrt 3-1|}\\=\dfrac{1}{\sqrt 3-1}\\=\dfrac{1+\sqrt3 }{2}\)
Vậy \(A=\dfrac{1+\sqrt 3}{2}\) tại \(x=4-2\sqrt 3\)
b/ \( B=(x+9).A-5\\=\dfrac{x+9}{\sqrt x}-5\\=\dfrac{x-5\sqrt x+9}{\sqrt x}\\=\sqrt x-5+\dfrac{9}{\sqrt x}\\=(\sqrt x+\dfrac{9}{\sqrt x})-5\\→B≥2\sqrt{\sqrt x.\dfrac{9}{\sqrt x}}-5\\→B≥1\\→\min B=1\)
\(→\) Dấu "=" xảy ra khi \(\sqrt x=\dfrac{9}{\sqrt x}\)
\(↔x=9(TM)\)
Vậy \(\min B=1\) khi \(x=9\)
