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

Решите задачу:

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

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