Question

FUNDA!!
Compara
radical din 17+radical din6 cu radical din 10+radical din15
CU EXPLICATIE PLSS

Answer 1

   
[tex] \sqrt{17} + \sqrt{6} ~\approx ~ \sqrt{10} +\sqrt{15} ~~\Big|~~ \text{Ridicam la patrat.}\\ \left(\sqrt{17} + \sqrt{6}~\right)^2 ~\approx ~ \left(\sqrt{10} +\sqrt{15} ~\right)^2\\ 289 +2 \sqrt{17}\sqrt{6}+36 ~\approx ~ 100+2\sqrt{10}\sqrt{15}+225 \\ 289+36 +2\sqrt{17 \times 6} ~\approx ~ 100+225+2\sqrt{10\times 15} \\ 325 +2\sqrt{17 \times 6} ~\approx ~ 325+2\sqrt{10\times 15} ~~\Big|~~ -325\\ 2\sqrt{17 \times 6} ~\approx ~ 2\sqrt{10\times 15} ~~\Big|~~ : 2 [/tex]


$\sqrt{17 \times 6} ~\approx ~ \sqrt{10\times 15} ~~\Big|~~ \text{Ridicam la patrat.}\\ 17 \times 6 ~\approx ~10\times 15\\ 102~\ \textless \ ~150\\ \Longrightarrow ~~\boxed{\bf \sqrt{17} + \sqrt{6} ~\ \textless \ ~ \sqrt{10} +\sqrt{15} }\\\\ \texttt{Din rezolvarea asta putem trage concluzia urmatoare:}\\\\ \boxed{\bf a\times b \ \textless \ c\times d ~~\Longrightarrow~~ \sqrt{a}+\sqrt{b} \ \textless \ \sqrt{c}+\sqrt{d}}$

$\text{Daca: }~~a+b \leq c+d$