Đáp án:
$1) B=\dfrac{2x}{1-x}\\ 2)\\ A=6\sqrt{3}+2.$
Giải thích các bước giải:
$1)\\ B=\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}-2(x \ge 0; x \ne 1)\\ =\dfrac{1+\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}+\dfrac{1-\sqrt{x}}{(1+\sqrt{x})(1-\sqrt{x})}-2\\ =\dfrac{1+\sqrt{x}}{1-x}+\dfrac{1+\sqrt{x}}{1-x}-2\\ =\dfrac{2}{1-x}-2\\ =\dfrac{2-2(1-x)}{1-x}\\ =\dfrac{2x}{1-x}\\ 2)\\ A=(5-\sqrt{11})(5+\sqrt{11})-(3-\sqrt{3})^2\\ =25-11-(9-6\sqrt{3}+3)\\ =14-(12-6\sqrt{3})\\ =6\sqrt{3}+2.$
