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[tex]e)\ \ \frac{2}{3}\sqrt{3}(5\sqrt{3}-4\sqrt{27}-12)=\\\\=\frac{2}{3}\sqrt{3}\cdot5\sqrt{3}-\frac{2}{3}\sqrt{3}\cdot4\sqrt{27}-\frac{2}{3}\sqrt{3}\cdot12=\frac{10}{3}\sqrt{9}-\frac{8}{3}\sqrt{81}-8\sqrt{3}=\\\\=\frac{10}{\not3}\cdot\not3-\frac{8}{\not3_{1}}\cdot\not9^3-8\sqrt{3}=10-24-8\sqrt{3}=-14-8\sqrt{3}[/tex]
[tex]f)\ \ -0,1\sqrt{15}(\sqrt{15}-2\sqrt{5}+20)=-0,1\sqrt{15}\cdot\sqrt{15}-0,1\sqrt{15}\cdot(-2\sqrt{5})-0,1\sqrt{15}\cdot20=\\\\=-0,1\cdot15+0,2\sqrt{75}-2\sqrt{15}=-1,5+0,2\cdot\sqrt{25\cdot3}-2\sqrt{15}=\\\\=-1,5+0,2\cdot5\sqrt{3}-2\sqrt{15}=-1,5+\sqrt{3}-2\sqrt{15}[/tex]
[tex]g)\ \ (5-2\sqrt{7})(5+2\sqrt{7})=25+10\sqrt{7}-10\sqrt{7}-4\cdot7=25-28=-3\\\\lub\\\\(5-2\sqrt{7})(5+2\sqrt{7})=5^2-(2\sqrt{7})^2=25-4\cdot7=25-28=-3\\\\\\h)\ \ (3-2\sqrt{2})(5+\sqrt{2})=15+3\sqrt{2}-10\sqrt{2}-2\cdot2=15-7\sqrt{2}-4=11-7\sqrt{2}[/tex]