Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 23:\ a.\ 20;\ b.\ 15;\ c.\ 26;\ d.\ 18\\ Bài\ 24:\ a.\ 60;\ b.\ 60;\ c.\ 24;\ d.\ 6\\ Bài\ 25:a.\ 6;\ b.\ 12;\ c.\ 108;\ d.\ 128\\ Bài\ 27:a.\ \frac{1}{\sqrt{2}} ;\ b.\ 1+\sqrt{2} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 23:\\ a.\ \sqrt{10} .\sqrt{40} =\sqrt{10.40} =\sqrt{400} =20\\ b.\ \sqrt{5} .\sqrt{45} =\sqrt{5.45} =\sqrt{225} =15\\ c.\sqrt{52} .\sqrt{13} =\ \sqrt{52.13} =\sqrt{676} =26\\ d.\ \sqrt{2} .\sqrt{162} =\sqrt{2.162} =\sqrt{324} =18\\ Bài\ 24:\\ a.\ \sqrt{45.80} =\sqrt{3600} =60\\ b.\ \sqrt{75.48} =\sqrt{3600} =60\\ c.\ \sqrt{90.6,4} =\sqrt{576} =24\\ d.\ \sqrt{2,5.14,4} =\sqrt{36} =6\\ Bài\ 25:\\ a.\ \sqrt{6,8^{2} -3,2^{2}} =\sqrt{( 6,8+3,2)( 6,8-3,2)}\\ =\sqrt{10.3,6} =\sqrt{36} =6\\ b.\ \sqrt{21,8^{2} -18,2^{2}} =\sqrt{( 21,8-18,2)( 21,8+18,2)}\\ =\sqrt{40.3,6} =\sqrt{144} =12\\ c.\ \sqrt{117,5^{2} -26,5^{2} -1440}\\ =\sqrt{( 117,5-26,5)( 117,5+26,5) -1440}\\ =\sqrt{91.144-1440} =\sqrt{144.( 91-10)} =\sqrt{144.81} =12.9=108\\ d.\ \sqrt{146,5^{2} -109,5^{2} +27.256}\\ =\sqrt{( 146,5-109,5)( 146,5+109,5) +27.256}\\ =\sqrt{37.256+27.256} =\sqrt{256.( 37+27)} =\sqrt{256.64} =16.8=128\\ Bài\ 26:\\ a.\ \sqrt{9-\sqrt{17}} .\sqrt{9+\sqrt{17}} =\sqrt{\left( 9-\sqrt{17}\right) .\left( 9+\sqrt{17}\right)}\\ =\sqrt{9^{2} -\left(\sqrt{17}\right)^{2}} =\sqrt{81-17} =\sqrt{64} =8\Rightarrow đpcm\\ b.\ 2\sqrt{2}\left(\sqrt{3} -2\right) +\left( 1+2\sqrt{2}\right)^{2} -2\sqrt{6}\\ =2\sqrt{6} -4+1+4+8-2\sqrt{6} =9\Rightarrow đpcm\\ Bài\ 27:\\ a.\ \frac{\sqrt{6} +\sqrt{14}}{2\sqrt{3} +\sqrt{28}} =\frac{\sqrt{2} .\left(\sqrt{3} +\sqrt{7}\right)}{2.\left(\sqrt{3} +\sqrt{7}\right)} =\frac{\sqrt{2}}{2} =\frac{1}{\sqrt{2}}\\ b.\ \frac{\sqrt{2} +\sqrt{3} +\sqrt{6} +\sqrt{8} +\sqrt{16}}{\sqrt{2} +\sqrt{3} +\sqrt{4}} =\frac{\sqrt{2} +\sqrt{3} +\sqrt{6} +\sqrt{8} +\sqrt{4} +\sqrt{4}}{\sqrt{2} +\sqrt{3} +\sqrt{4}}\\ =\frac{\left(\sqrt{2} +\sqrt{3} +\sqrt{4}\right) +\sqrt{2} .\left(\sqrt{2} +\sqrt{3} +\sqrt{4}\right)}{\sqrt{2} +\sqrt{3} +\sqrt{4}} =1+\sqrt{2} \end{array}$
