Question

$\sqrt{1}7$ – $\sqrt{1}2$ and $\sqrt{1}1$ – $\sqrt{6}$. Which is greater?

Answer 1

$\begin{array}{cc}\sqrt{11}-\sqrt{12}& \sqrt{11}-\sqrt{6}\\ =\frac{\sqrt{17}-\sqrt{12}}{1}\times \frac{\sqrt{17}+\sqrt{12}}{\sqrt{17}+\sqrt{12}}& =\frac{\sqrt{1}-\sqrt{6}}{1}\times \frac{\sqrt{11}+\sqrt{6}}{\sqrt{11}+\sqrt{6}}\\ & \frac{11-6}{\sqrt{11}+\sqrt{6}}\\ =\frac{17-12}{\sqrt{11}+\sqrt{12}}& \frac{5}{\sqrt{11}+\sqrt{6}}\\ =\frac{5}{\sqrt{17}+\sqrt{12}}& 17>11\Rightarrow \sqrt{17}>\sqrt{11}\\ & 12>6\Rightarrow \sqrt{12}>\sqrt{6}\\ & \Rightarrow \sqrt{17}+\sqrt{12}>\sqrt{11}+\sqrt{6}\\ \frac{5}{\sqrt{17}+\sqrt{12}}<\frac{5}{\sqrt{11}+\sqrt{6}}& (\because Infroctionshavingsamenumetator\\ & \&differentdinomintors,the\\ & fractionwithsmalldenomintorislongerinvalue)\\ \Rightarrow \sqrt{17}-\sqrt{12}<\sqrt{11}-\sqrt{6}& \end{array}$