Đáp án:

$1)\\ A=\sqrt{3}-2\\ B=1\\ C=1\\ D=1\\ E=-\sqrt{5}+15\sqrt{2}\\ 3)\\ A=-\dfrac{1}{x}\\ B=\dfrac{x-9}{(\sqrt{x}-2)(x-3)}\\ C=\sqrt{x}-1$

Giải thích các bước giải:

$1)\\ A=\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\dfrac{2+\sqrt{8}}{1+\sqrt{2}}\\ =\dfrac{\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}-\dfrac{2+\sqrt{4.2}}{1+\sqrt{2}}\\ =\sqrt{3}-\dfrac{2+2\sqrt{2}}{1+\sqrt{2}}\\ =\sqrt{3}-\dfrac{2(1+\sqrt{2})}{1+\sqrt{2}}\\ =\sqrt{3}-2\\ B=\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right).\sqrt{6}\\ =\sqrt{\dfrac{3}{2}}.\sqrt{6}-\sqrt{\dfrac{2}{3}}.\sqrt{6}\\ =\sqrt{\dfrac{3.6}{2}}-\sqrt{\dfrac{2.6}{3}}\\ =\sqrt{\dfrac{18}{2}}-\sqrt{\dfrac{12}{3}}\\ =\sqrt{9}-\sqrt{4}\\ =3-2\\ =1\\ C=\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\\ =\left(2+\dfrac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}\right)\left(2-\dfrac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1}\right)\\ =\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\\ =4-3\\ =1\\ D=3\sqrt{8}-\sqrt{50}-\sqrt{(\sqrt{2}-1)^2}\\ =3\sqrt{4.2}-\sqrt{25.2}-|\sqrt{2}-1|\\ =3.2\sqrt{2}-5\sqrt{2}-(\sqrt{2}-1)\\ =6\sqrt{2}-5\sqrt{2}-\sqrt{2}+1\\ =1\\ E=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\\ =\sqrt{4.5}-\sqrt{9.5}+3\sqrt{9.2}+\sqrt{36.2}\\ =2\sqrt{5}-3\sqrt{5}+3.3\sqrt{2}+6\sqrt{2}\\ =2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\\ =-\sqrt{5}+15\sqrt{2}\\ 3)\\ A=\dfrac{2}{x-1}\sqrt{\dfrac{x^2-2x+1}{4x^2}} \ \ \ \ 0<x<1\\ =\dfrac{2}{x-1}\sqrt{\dfrac{(x-1)^2}{(2x)^2}} \\ =\dfrac{2}{x-1}.\dfrac{|x-1|}{|2x|}\\ =\dfrac{2}{x-1}.\dfrac{1-x}{2x}\\ =-\dfrac{1}{x}\\ B=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-3}{\sqrt{x}-3} \ \ \ \ x \ge 0, x \ne 4;x \ne 3; x \ne 9\\ =\left(\dfrac{3\sqrt{x}+6}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right).\dfrac{\sqrt{x}-3}{x-3}\\ =\left(\dfrac{3\sqrt{x}+6}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\right).\dfrac{\sqrt{x}-3}{x-3}\\ =\dfrac{3\sqrt{x}+6+\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{\sqrt{x}-3}{x-3}\\ =\dfrac{x+5\sqrt{x}+6}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{\sqrt{x}-3}{x-3}\\ =\dfrac{(\sqrt{x}+2)(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{\sqrt{x}-3}{x-3}\\ =\dfrac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(x-3)}\\ =\dfrac{x-9}{(\sqrt{x}-2)(x-3)}\\ C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1} \ \ \ \ x>0; x\ne 1\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right).\dfrac{x-1}{\sqrt{x}+1} \\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}+1}\\  =\dfrac{\sqrt{x}-1}{\sqrt{x}-1}.(\sqrt{x}-1)\\ =\sqrt{x}-1$