Question

DAU 20 PCT!!! :(3-2√2)√(mare)17+12√2+√(mare)7+4√3(√3-2)

Answer 1

(3-2√2)$\sqrt{17+12 \sqrt{2} }$+(√3-2)$\sqrt{7+4 \sqrt{3} }$
Luam pe rand:
17+12√2=(3+2√2)²
7+4√3=(√3+2)²
(3-2√2)$\sqrt{(3+2 \sqrt{2} )^2}$+(√3-2)$\sqrt{( \sqrt{3} +2)^2}$=
=(3-2√2)|3+2√2|+(√3-2)|√3+2|=
=(3-2√2)(3+2√2)+(√3-2)(√3+2)=
=9-8+3-4=0