Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 53.D\\ 54.\ A\\ 55.C\\ 56.A\\ 57.C\\ 58.A\\ 59.D\\ 60.B\\ 61.C\\ 62.B\\ 63.C\\ 64.B\\ 65.A\\ 66.D \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 53.D\\ \left( 1+\frac{3-\sqrt{3}}{\sqrt{3} -1}\right) .\left(\frac{3+\sqrt{3}}{\sqrt{3} +1} -1\right) =\frac{\sqrt{3} -1+3-\sqrt{3}}{\sqrt{3} -1} .\frac{3-\sqrt{3} -\sqrt{3} -1}{\sqrt{3} +1}\\ =\frac{2}{\sqrt{3} -1} .\frac{2-2\sqrt{3}}{\sqrt{3} +1} =2\\ 54.\ A\\ \sqrt{9} =\sqrt{3^{2}} =3\\ 55.C\\ \sqrt{4-3x} \ xác\ định\ \Leftrightarrow 4-3x\geqslant 0\Leftrightarrow x\leqslant \frac{4}{3}\\ 56.A\\ P=|1-\sqrt{3} |-|1+\sqrt{3} |=\sqrt{3} -1-1-\sqrt{3} =-2\\ 57.C\\ 2-\sqrt{\left(\sqrt{3} -2\right)^{2}} =2-|\sqrt{3} -2|=2-2+\sqrt{3} =\sqrt{3}\\ 58.A\\ \frac{y}{x}\sqrt{\frac{x^{2}}{y^{4}}} =\frac{y}{x} .\frac{x}{y^{2}} =\frac{1}{y}\\ 59.D\\ \sqrt{3} x=\sqrt{12} \Leftrightarrow x=\frac{\sqrt{12}}{\sqrt{3}} =\sqrt{4} =2\\ 60.B\\ \sqrt{3x-5} \ xác\ \ định\ \Leftrightarrow 3x-5\geqslant 0\Leftrightarrow x\geqslant \frac{5}{3}\\ 61.C\\ B=3.( -3) -2.2=-5\\ 62.B\\ \sqrt{x-2} =3\Leftrightarrow x-2=9\Leftrightarrow x=11\ ( TM)\\ 63.C\\ P( x) \ xác\ \ định\ \ \Leftrightarrow 2013-2014x\geqslant 0\Leftrightarrow x\leqslant \frac{2013}{2014}\\ 64.B\\ A=|\sqrt{5} -3|+|\sqrt{5} -2|-1=3-\sqrt{5} +\sqrt{5} -2-1=1-1=0\\ 65.A\\ A\ xác\ \ định\ \Leftrightarrow 2014-2015x\geqslant 0\Leftrightarrow x\leqslant \frac{2014}{2015}\\ 66.D\\ x\sqrt{\frac{1}{x^{2}}} =x.|\frac{1}{x} |=\frac{x}{-x} \ ( do\ x< 0) =-1 \end{array}$