Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a\ .\ 15\sqrt{2}\\ b.\ 1\\ c.\ -3\\ Bài\ 2.\\ a.\ S=\left\{0;\frac{2}{3}\right\}\\ b.\ S=\{9\} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a\ .\ 6\sqrt{2} -3\sqrt{2} +15\sqrt{2} -3\sqrt{2} =15\sqrt{2}\\ b.\ \sqrt{17+12\sqrt{2}}\left( 3-\sqrt{8}\right) =\sqrt{\left( 3+2\sqrt{2}\right)^{2}}\left( 3-\sqrt{8}\right)\\ =\left( 3+2\sqrt{2}\right)\left( 3-2\sqrt{2}\right) =9-8=1\\ c.\ \frac{\sqrt{10} -3}{10-9} +\frac{\sqrt{10}\left(\sqrt{5} -\sqrt{2}\right)}{\sqrt{2} -\sqrt{5}} =\sqrt{10} -3-\sqrt{10} =-3\\ Bài\ 2.\\ a.\ \sqrt{( 6x-2)^{2}} =2\\ \Leftrightarrow 2|3x-1|=2\\ TH\ 1:\ x\geqslant \frac{1}{3}\\ \Leftrightarrow 3x-1=1\\ \Leftrightarrow x=\frac{2}{3} \ ( TM)\\ TH2:\ x< \frac{1}{3}\\ \Leftrightarrow 1-3x=1\\ \Leftrightarrow x=0\ ( TM)\\ Vậy\ S=\left\{0;\frac{2}{3}\right\}\\ b.\ ĐK:\ x\geqslant 5\\ \Leftrightarrow 2\sqrt{x-5} -\sqrt{x-5} =10-4\sqrt{x-5}\\ \Leftrightarrow 5\sqrt{x-5} =10\\ \Leftrightarrow \sqrt{x-5} =2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\ ( TM)\\ Vậy\ S=\{9\} \end{array}$
