ĐKXĐ: \(x\ne 1,x\ge 0\)
\(B=(\dfrac{1}{\sqrt x-1}-\dfrac{1}{x\sqrt x -1}).\dfrac{3\sqrt x-3}{x+\sqrt x}\\=(\dfrac{1}{\sqrt x-1}-\dfrac{1}{(\sqrt x)^2-1}).\dfrac{3\sqrt x-3}{x+\sqrt x}\\=(\dfrac{x+\sqrt x+1}{(\sqrt x-1)(x+\sqrt x+1)}-\dfrac{1}{(\sqrt x-1)(x+\sqrt x+1)}).\dfrac{3\sqrt x-3}{x+\sqrt x}\\=\dfrac{x+\sqrt x+1-1}{(\sqrt x-1)(x+\sqrt x+1)}.\dfrac{3(\sqrt x-1)}{x+\sqrt x}\\=\dfrac{x+\sqrt x}{(\sqrt x-1)(x+\sqrt x+1)}.\dfrac{3(\sqrt x-1)}{x+\sqrt x}\\=\dfrac{3}{x+\sqrt x+1}\)
\(x=4+2\sqrt 3\\=3+2\sqrt 3+1\\=(\sqrt 3+1)^2\)
Thay \(x=(\sqrt 3+1)^2(TM)\) vào biểu thức \(B\)
\(B=\dfrac{3}{(\sqrt 3+1)^2+\sqrt{(\sqrt 3+1)^2}+1}\\=\dfrac{3}{4+2\sqrt 3+|\sqrt 3+1|+1}\\=\dfrac{3}{4+2\sqrt 3+\sqrt 3+1+1}\\=\dfrac{3}{6+3\sqrt 3}\\=\dfrac{3}{3(2+\sqrt 3)}\\=\dfrac{1}{2+\sqrt 3}\\=2-\sqrt 3\)
Vậy \(B=2-\sqrt 3\) tại \(x=4+2\sqrt 3\)
