Odpowiedź :
a)
[tex]3\sqrt{5} + \sqrt{32} -2\sqrt{18}=3\sqrt{5} +\sqrt{16\times 2} -2\sqrt{9 \times 2} =3\sqrt{5} + 4\sqrt{2} - 6\sqrt{2} = 3\sqrt{5} -2\sqrt{2}[/tex]
b)
[tex]\sqrt{3} ( \sqrt{3} - 2 \sqrt{27} + \sqrt{48}) = \sqrt{3} ( \sqrt{3} - 2 \sqrt{9 \times 3} + \sqrt{16 \times 3} ) = \sqrt{3} ( \sqrt{3} - 6 \sqrt{3} + 4 \sqrt{3} ) = \sqrt{3} ( - \sqrt{3} ) = - \sqrt{9} = - 3[/tex]
c)
[tex] \sqrt{5} ( \sqrt{5} - 5 \sqrt{125} + \sqrt{500} ) = \sqrt{5} ( \sqrt{5} - 5 \sqrt{25 \times 5} + \sqrt{100 \times 5} ) = \sqrt{5} ( \sqrt{5} - 25 \sqrt{5} + 10 \sqrt{5} ) = \sqrt{5} ( - 14 \sqrt{5} ) = - \sqrt{5} \times 14 \sqrt{5} = - 5 \times 14 = - 70[/tex]
Myślę że pomogłem ;)
Odpowiedź:
[tex]3\sqrt{8}+\sqrt{32}-2\sqrt{18}=3\sqrt{4\cdot2}+\sqrt{16\cdot2}-2\sqrt{9\cdot2}=3\cdot2\sqrt{2}+4\sqrt{2}-2\cdot3\sqrt{2}=\\\\=6\sqrt{2}+4\sqrt{2}-6\sqrt{2}=4\sqrt{2}\\\\\\\sqrt{3}(\sqrt{3}-2\sqrt{27}+\sqrt{48})=\sqrt{9}-2\sqrt{81}+\sqrt{144}=3-2\cdot9+12=3-18+12=\\\\=15-18=-3[/tex]
[tex]\sqrt{5}(\sqrt{5}-5\sqrt{125}+\sqrt{500})=\sqrt{5}(\sqrt{5}-5\sqrt{25\cdot5}+\sqrt{100\cdot5})=\\\\=\sqrt{5}(\sqrt{5}-5\cdot5\sqrt{5}+10\sqrt{5})=\sqrt{5}(\sqrt{5}-25\sqrt{5}+10\sqrt{5})=\sqrt{5}\cdot(-14\sqrt{5})=\\\\=-5\cdot14=-70\\\\\\lub\\\\\sqrt{5}(\sqrt{5}-5\sqrt{125}+\sqrt{500})=\sqrt{25}-5\sqrt{625}+\sqrt{2500}=5-5\cdot25+50=\\\\=5-125+50=55-125=-70[/tex]