Odpowiedź:
[tex]\sqrt{7^2}+\sqrt{6}-(\sqrt{6})^2=7+\sqrt{6}-6=1+\sqrt{6}\\\\\\\sqrt[3]{12^3}-(\sqrt{17})^2-\sqrt{5}=12-17-\sqrt{5}=-5-\sqrt{5}\\\\\\3\sqrt{5^2}+2\sqrt{5}-2\sqrt[3]{(-5)^3}=3\cdot5+2\sqrt{5}-2\cdot(-5)=15+2\sqrt{5}+10=25+2\sqrt{5}\\\\\\-\sqrt[3]{5^3}-\sqrt[3]{7}-\sqrt[3]{-7}=-5-\sqrt[3]{7}-(-\sqrt[3]{7})=-5-\sqrt[3]{7}+\sqrt[3]{7}=-5\\\\\\5\sqrt{2}+3(1-2\sqrt{2})=5\sqrt{2}+3-6\sqrt{2}=-\sqrt{2}+3[/tex]
[tex]3\sqrt{5}+2(\sqrt{5}-3)=3\sqrt{5}+2\sqrt{5}-6=5\sqrt{5}-6\\\\\\(3+\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+\sqrt{2}+\sqrt{4}=3+3\sqrt{2}+\sqrt{2}+2=5+4\sqrt{2}[/tex]
