Question

greater number along √17-√12 and √11-√6

Answer 1

$\sqrt{17 } - \sqrt{12} \: and \: \sqrt{11} - \sqrt{16} \\ frist........ \\ ({ \sqrt{17} - \sqrt{12} })^{2} \\ = 17 + 12 - 2 \sqrt{204} \\ = 29 - 2 \sqrt{204} \\ \\ then...... \\ \sqrt{11} - \sqrt{16} \\ = { \(sqrt{11} - \sqrt{16} \\ = { \(sqrt{11} - \sqrt{16} })^{2} \\ = 11 + 16 - 2 \sqrt{176} \\ = 27 - 2 \sqrt{176} \\ \\ therefore...... \sqrt{17} - \sqrt{12} > \sqrt{11} - \sqrt{16}$