Math, asked by ayamakhoukhi8, 1 year ago

prove that root (x square+1)-x>0

Answers

Answered by chopraneetu
0

 \sqrt{ {x}^{2} }  = x \\  \sqrt{ {x}^{2}  + 1}  > x \\ \:  \sqrt{ {x}^{2}  + 1}   - x > 0
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