Đáp án:
$A = - \sqrt{2}$
$B = -2-\dfrac{2}{x-1}$
Giải thích các bước giải:
$\begin{array}{l}A=\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{5}}}\\=\sqrt{\dfrac{(3\sqrt{3}-4)(2\sqrt{3}-1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)}}-\sqrt{\dfrac{(\sqrt{3}+4)(5-2\sqrt{5})}{(5-2\sqrt{5})(5+2\sqrt{5})}}\\=\sqrt{\dfrac{22-11\sqrt{3}}{11}}-\sqrt{\dfrac{26+13\sqrt{3}}{13}}\\=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\\=\dfrac{2(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}})}{2}\\=\dfrac{\sqrt{2}}{2}.(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}})\\=\dfrac{\sqrt{2}}{2}.[\sqrt{3} - 1 - (\sqrt{3} + 1)]\\=-\sqrt{2}\\\\B=\left(\dfrac{\sqrt{x}+1}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).(x+\sqrt{x})\\=\left[\dfrac{(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}\right].[\sqrt{x}(\sqrt{x}+1)]\\=\dfrac{x + \sqrt{x} - 2 - (x-\sqrt{x}-2)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}.\sqrt{x}(\sqrt{x}+1)\\=\dfrac{2\sqrt{x}.\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}\\=\dfrac{2x}{x-1}\\=2+\dfrac{2}{x-1}\end{array}$
