Question

Cum pot calcula radical mare din 3+2radical din 2. Primul radical se afla si peste 2radical2

Answer 1

[tex] \sqrt{3+2 \sqrt{2} } = \sqrt{1+2\sqrt{2} +2} = \sqrt{1^2+2*1*\sqrt{2} +(\sqrt{2} )^2} \ \textless \ =\ \textgreater \ \\ \sqrt{3+2 \sqrt{2} } = \sqrt{(1+\sqrt{2} )^2} =|1+\sqrt{2} | \\ 1+\sqrt{2} \ \textgreater \ 0=\ \textgreater \ |1+\sqrt{2} |=1+\sqrt{2} =\ \textgreater \ \\ \sqrt{3+2 \sqrt{2} } =1+ \sqrt{2} [/tex]

Sper sa te ajute.

Answer 2

$2 \sqrt{2} = \sqrt{2*2*2} = \sqrt{8}$

se aplica formula radicalilor compusi
$\sqrt{a+ \sqrt{b} } = \sqrt{ \frac{a+ \sqrt{a^{2}-b}}{2}}+ \sqrt{ \frac{a- \sqrt{a^{2}-b}}{2}}\\ \\ \sqrt{3+2 \sqrt{2} } = \sqrt{3+ \sqrt{8} } = \sqrt{ \frac{3+ \sqrt{3^{2}-8}}{2}}+ \sqrt{ \frac{3- \sqrt{3^{2}-8}}{2}} \\ \\ = \sqrt{ \frac{3+1}{2}}+ \sqrt{ \frac{3- 1}{2}} \\ \\ = \sqrt{2} + \sqrt{1} \\ \\ =1+ \sqrt{2}$