Question

b) (radical din 7 -radical din 5) la 2?

Answer 1

${( \sqrt{7} - \sqrt{5} ) }^{2} = {( \sqrt{7}) }^{2} - 2 \times \sqrt{7} \times \sqrt{5} + {( \sqrt{5} )}^{2}$

$= 7 - 2 \sqrt{7 \times 5} + 5$

$= 7 - 2 \sqrt{35} + 5$

$= 12 - 2 \sqrt{35}$

$= 2(6 - \sqrt{35} )$

$Formul\check{a}\::{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$