[tex]\boxed{a)\ (1+\sqrt2)(2+\sqrt2)=4+3\sqrt2}\\\boxed{b)\ (2+\sqrt3)(5-\sqrt3)=7+3\sqrt3}\\\boxed{c)\ (4-2\sqrt5)(2-3\sqrt5)=38-16\sqrt5}\\\boxed{d)\ (\sqrt3+\sqrt2)(2\sqrt3-\sqrt2)=4+\sqrt6}\\\boxed{e)\ (\sqrt2+\sqrt6)(\sqrt3+\sqrt2)=2+3\sqrt2+2\sqrt3+\sqrt6}\\\boxed{f)\ (2\sqrt5-3\sqrt2)(\sqrt{10}-3\sqrt5)=10\sqrt2+9\sqrt{10}-6\sqrt5-30}[/tex]
Skorzystamy z rozdzielności mnożenia względem dodawania/odejmowania:
(a + b)(c + d) = ab + ad + bc + bd
oraz z twierdzeń
√a · √a = a dla a ≥ 0
√(a · b) = √a · √b dla a,b ≥ 0
[tex]a)\ (1+\sqrt2)(2+\sqrt2)=1\cdot2+1\cdot\sqrt2+\sqrt2\cdot2+\sqrt2\cdot\sqrt2\\\\=2+\sqrt2+2\sqrt2+2=\boxed{4+3\sqrt2}\\\\b)\ (2+\sqrt3)(5-\sqrt3)=2\cdot5-2\cdot\sqrt3+\sqrt3\cdot5-\sqrt3\cdot\sqrt3\\\\=10-2\sqrt3+5\sqrt3-3=\boxed{7+3\sqrt3}\\\\c)\ (4-2\sqrt5)(2-3\sqrt5)=4\cdot2-4\cdot3\sqrt5-2\sqrt5\cdot2+2\sqrt5\cdot3\sqrt5\\\\=8-12\sqrt5-4\sqrt5+6\cdot5=\boxed{38-16\sqrt5}\\\\d)\ (\sqrt3+\sqrt2)(2\sqrt3-\sqrt2)=\sqrt3\cdot2\sqrt3-\sqrt3\cdot\sqrt2+\sqrt2\cdot2\sqrt3-\sqrt2\cdot\sqrt2\\\\=2\cdot3-\sqrt6+2\sqrt6-2=\boxed{4+\sqrt6}\\\\e)\ (\sqrt2+\sqrt6)(\sqrt3+\sqrt2)=\sqrt2\cdot\sqrt3+\sqrt2\cdot\sqrt2+\sqrt6\cdot\sqrt3+\sqrt6\cdot\sqrt2\\\\=\sqrt6+2+\sqrt{18}+\sqrt{12}=\sqrt6+2+\sqrt{9\cdot2}+\sqrt{4\cdot3}\\\\=\sqrt6+2+\sqrt9\cdot\sqrt2+\sqrt4\cdot\sqrt3=\boxed{2+3\sqrt2+2\sqrt3+\sqrt6}\\\\f)\ (2\sqrt5-3\sqrt2)(\sqrt{10}-3\sqrt5)=2\sqrt5\cdot\sqrt{10}-2\sqrt5\cdot3\sqrt5-3\sqrt2\cdot\sqrt{10}+3\sqrt2\cdot3\sqrt5\\\\=2\sqrt{50}-6\cdot5-3\sqrt{20}+9\sqrt{10}=2\sqrt{25\cdot2}-30-3\sqrt{4\cdot5}+9\sqrt{10}\\\\=2\cdot\sqrt{25}\cdot\sqrt2-30-3\cdot\sqrt4\cdot\sqrt5+9\sqrt{10}=2\cdot5\cdot\sqrt2-30-3\cdot2\cdot\sqrt5+9\sqrt{10}\\\\=\boxed{10\sqrt2+9\sqrt{10}-6\sqrt5-30}[/tex]