(3-2√2)[tex] \sqrt{17+12 \sqrt{2} } [/tex]+(√3-2)[tex] \sqrt{7+4 \sqrt{3} } [/tex]
Luam pe rand:
17+12√2=(3+2√2)²
7+4√3=(√3+2)²
(3-2√2)[tex] \sqrt{(3+2 \sqrt{2} )^2} [/tex]+(√3-2)[tex] \sqrt{( \sqrt{3} +2)^2} [/tex]=
=(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
