Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.\\ a.\ A=1\\ b.\ K=\frac{\sqrt{5}}{2}\\ c.\ L=-2\sqrt{2} -1\\ 3.\\ a.\ 3\\ b.\ \sqrt{7}\\ c.\ A=\sqrt{3} -2\\ d.\ \frac{4\sqrt{3}}{3} ;\ \frac{5+\sqrt{5}}{4}\\ e.\ D=1\\ f.\ E=1\\ h.\ H=\sqrt{2}\\ k.\ K=\frac{\sqrt{5}}{2}\\ m.\ L=1 \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.\\ a.\ A=\sqrt{\left(\sqrt{6} +1\right)^{2}} -\sqrt{6} =\sqrt{6} +1-\sqrt{6} =1\\ b.\ K=\frac{1}{4}\frac{4}{3-\sqrt{5}} -\frac{1}{4}\frac{4}{\sqrt{5} +3} =\frac{1}{4}\left( 3+\sqrt{5} -3+\sqrt{5}\right) =\frac{\sqrt{5}}{2}\\ c.\ L=9\sqrt{2} -10\sqrt{2} -\sqrt{\left(\sqrt{2} +1\right)^{2}} =-\sqrt{2} -\sqrt{2} -1=-2\sqrt{2} -1\\ 3.\\ a.\ 2+\sqrt{3} +2−\sqrt{3} -\left( 2+\sqrt{3}\right)\left( 2−\sqrt{3}\right) =4-4+3=3\\ b.\ \frac{1}{2}\left(\frac{2}{3-\sqrt{7}} -\frac{2}{3+\sqrt{7}}\right) =\frac{1}{2}\left( 3+\sqrt{7} -3+\sqrt{7}\right) =\sqrt{7}\\ c.\ A=\frac{\sqrt{3}\left( 1-\sqrt{2}\right)}{1-\sqrt{2}} -\frac{2\left( 1+\sqrt{2}\right)}{1+\sqrt{2}} =\sqrt{3} -2\\ d.\ \frac{4}{\sqrt{3}} =\frac{4\sqrt{3}}{3} ;\ \frac{\sqrt{5}}{\sqrt{5} -1} =\frac{\sqrt{5}\left(\sqrt{5} +1\right)}{5-1} =\frac{5+\sqrt{5}}{4}\\ e.\ D=\frac{\sqrt{3}\sqrt{6}}{\sqrt{2}} -\frac{\sqrt{2}\sqrt{6}}{\sqrt{3}} =\sqrt{9} -\sqrt{4} =3-2=1\\ f.\ E=\left( 2+\sqrt{3}\right)\left( 2-\sqrt{3}\right) =4-3=1\\ h.\ H=\sqrt{\left(\sqrt{3} +\sqrt{2}\right)^{2}} -\sqrt{3} =\sqrt{3} +\sqrt{2} -\sqrt{3} =\sqrt{2}\\ k.\ K=\frac{1}{4}\frac{4}{3-\sqrt{5}} -\frac{1}{4}\frac{4}{\sqrt{5} +3} =\frac{1}{4}\left( 3+\sqrt{5} -3+\sqrt{5}\right) =\frac{\sqrt{5}}{2}\\ m.\ L=6\sqrt{2} -5\sqrt{2} -\sqrt{2} +1=1 \end{array}$
