Question

(rad(5) +rad(3)-2)^2 - 2(rad(15)-2rad(5))+rad(48) +1
mă puteți ajuta să calculez asta,vă rog!
Dau coroană!

Answer 1

$(\sqrt{5}+\sqrt{3}-2)^2 - 2(\sqrt{15}-2\sqrt{5})+\sqrt{48}+1 =$

$=\left[(\sqrt{5}+\sqrt{3})-2\right]^2 - 2\sqrt{15}+4\sqrt{5}+\sqrt{16\cdot 3}+1$

$=(\sqrt{5}+\sqrt{3})^2-4(\sqrt{5}+\sqrt{3})+4-2\sqrt{15}+4\sqrt{5}+4\sqrt{3}+1$

$=(\sqrt{5}+\sqrt{3})^2-4(\sqrt{5}+\sqrt{3})+4(\sqrt{5}+\sqrt{3})-2\sqrt{15}+5$

$=5+2\cdot \sqrt{5}\cdot \sqrt{3}+3-2\sqrt{15}+5$

$=5+2\sqrt{15}+3-2\sqrt{15}+5$

$=5+3+5$

$=\mathbf{13}$