Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \ \frac{1}{7}\sqrt{51} \ < \frac{1}{9}\sqrt{150} \ \\ b.\ \sqrt{2017} -\sqrt{2016} \ < \ \sqrt{2016} -\sqrt{2015} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{1}{7}\sqrt{51} \ với\ \frac{1}{9}\sqrt{150}\\ Ta\ có:\ \frac{1}{7} .\sqrt{51} =\sqrt{\frac{51}{49}} =\sqrt{\frac{4131}{3969}}\\ \frac{1}{9} .\sqrt{150} =\sqrt{\frac{150}{81}} =\sqrt{\frac{7350}{3969}}\\ mà\ \frac{4131}{3969} < \frac{7350}{3969} \Leftrightarrow \sqrt{\frac{4131}{3969}} < \sqrt{\frac{7350}{3969}}\\ \ hay\ \frac{1}{7}\sqrt{51} \ < \frac{1}{9}\sqrt{150} \ \\ b.\ \sqrt{2017} -\sqrt{2016} \ với\ \sqrt{2016} -\sqrt{2015}\\ Ta\ có:\ \sqrt{2017} -\sqrt{2016} \ \\ =\frac{\left(\sqrt{2017} -\sqrt{2016} \ \right)\left(\sqrt{2017} +\sqrt{2016} \ \right)}{\sqrt{2017} +\sqrt{2016} \ }\\ =\frac{2017-2016}{\sqrt{2017} +\sqrt{2016} \ } =\frac{1}{\sqrt{2017} +\sqrt{2016} \ }\\ Tương\ tự:\ \sqrt{2016} -\sqrt{2015} =\frac{1}{\sqrt{2016} +\sqrt{2015}}\\ mà\sqrt{2017} +\sqrt{2016} >\ \sqrt{2016} +\sqrt{2015}\\ \Leftrightarrow \frac{1}{\sqrt{2017} +\sqrt{2016} \ } < \frac{1}{\sqrt{2016} +\sqrt{2015}}\\ hay\ \sqrt{2017} -\sqrt{2016} \ < \ \sqrt{2016} -\sqrt{2015} \end{array}$
