Question

[tex] \sqrt{} \sqrt{3}- \sqrt{8} =
[/tex]
$\sqrt{} \sqrt {9}+4 \sqrt{2} =$

Answer 1

Inainte de radical din 3 mai este un radical?

$\sqrt{a-\sqrt{b} } = \sqrt{ \frac{a+c}{2} } - \sqrt{ \frac{a-c}{2} }, c^{2} = a^{2} -b$\sqrt{3- \sqrt{8} } [/tex]
$c^{2} =9-8 => c=1$
[tex] \sqrt{ \frac{3+1}{2} } - \sqrt{ \frac{3-1}{2} }= \sqrt{2} -1 [/tex]
 

\sqrt{a+\sqrt{b} } = \sqrt{ \frac{a+c}{2} } + \sqrt{ \frac{a-c}{2} }, c^{2} = a^{2} -b
[tex] \sqrt{9+4 \sqrt{2} } [/tex]
$c ^{2} =81-32=> c=7$
$\sqrt{ \frac{9+7}{2} }+ \sqrt{ \frac{9-7}{2} } = 2 \sqrt{2} +1$