Question

(6-3√5)(6+3√5)+(3√5-2√20)*√5 pojak rewite zadanie)

Answer 1

$(6-3sqrt5)(6+3sqrt5)+(3sqrt5-2sqrt20)sqrt5=36-9cdot 5+3cdot 5-2sqrt{20cdot 5}=\\=36-45+15-2sqrt{4cdot 5cdot 5}=6-2cdot 10=-14$


$(a-b)(a+b)=a^2-b^2\\a=6,; b=3sqrt5\\a^2=36,; b^2=(3sqrt5)^2=3^2cdot (sqrt5)^2=9cdot 5=45\\a^2-b^2=36-45=-9\\(3sqrt5-2sqrt20)sqrt5=3sqrt5cdot sqrt5-2sqrt{20}cdot sqrt5=3cdot 5-2sqrt{20cdot 5}=\\=15-20=-5$

Answer 2

zdrastvuite nnnlll54 vy ne mogli by ob'yasnit' pochemu poluchilos 36-9*5 i v konce

Answer 3

Объяснила