Question

f:[0,infinit]--> R , f(x)=√x
1 supra f(1)+f(2) + 1 supra f(2)+f(3) + 1 supra f(3)+f(4)

Answer 1

f(1) = √1 = 1
f(2) = √2
f(3) = √3
f(4) = √4 = 2

Deci $\frac{1}{1+ \sqrt{2} } + \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+2 } =$
Rationalizam (aplificam prima fractie cu 1-√2, a doua fractie cu √2-√3, iar a treia fractie cu √3-2)
$= \frac{1- \sqrt{2} }{1-2} + \frac{ \sqrt{2} - \sqrt{3} }{2-3} + \frac{ \sqrt{3}-2 }{3-2} =$
Observam ca toti numitorii sunt -1 => putem sa scriem totul supra aceluiasi numitor.
$\frac{1- \sqrt{2}+ \sqrt{2}- \sqrt{3}+ \sqrt{3} -2 }{-1} = \frac{1-2}{-1} = \frac{-1}{-1} =1$