Задание 11 Параграф 17 ГДЗ Рабочая тетрадь 2 Мерзляк Полонский 8 класс (Алгебра)

Упростите выражение:

1. $$\sqrt{(82-1)}\cdot\sqrt{(82+1)}$$;
2. $$\sqrt{(58+\sqrt{33})}\cdot\sqrt{(58-\sqrt{33})}$$;
3. $$(\sqrt{2}+3)^2-(5-2\sqrt{2})^2$$;
4. $$(4+3\sqrt{6})^2+(4-3\sqrt{6})^2$$;
5. $$(16+6\sqrt{7})(3-\sqrt{7})^2$$;
6. $$(7-2\sqrt{10})(\sqrt{5}+\sqrt{2})^2$$;
7. $$(\sqrt{12-2\sqrt{11}}+\sqrt{12+2\sqrt{11}})^2$$;
8. $$(\sqrt{9+3\sqrt{5}}-\sqrt{9-3\sqrt{5}})^2$$.

1. $$\sqrt{\sqrt{82}-1}\cdot \sqrt{\sqrt{82}+1}=\sqrt{(\sqrt{82}-1)(\sqrt{82}+1)}=\sqrt{82-1}=\sqrt{81}=9.$$

2. $$\sqrt{\sqrt{58}+\sqrt{33}}\cdot \sqrt{\sqrt{58}-\sqrt{33}}=\sqrt{(\sqrt{58}+\sqrt{33})(\sqrt{58}-\sqrt{33})}=\sqrt{58-33}=\sqrt{25}=5.$$

3. $$(\sqrt{2}+3)^2-(5-2\sqrt{2})^2= (2+6\sqrt{2}+9)-(25-20\sqrt{2}+8)$$

   $$=11+6\sqrt{2}-33+20\sqrt{2}=26\sqrt{2}-22.$$

4. $$(4+3\sqrt{6})^2+(4-3\sqrt{6})^2$$

   $$=(16+24\sqrt{6}+54)+(16-24\sqrt{6}+54)=140.$$

5. $$(16+6\sqrt{7})(3-\sqrt{7})^2=(16+6\sqrt{7})(9-6\sqrt{7}+7)$$

   $$=(16+6\sqrt{7})(16-6\sqrt{7})=16^2-(6\sqrt{7})^2=256-252=4.$$

6. $$(7-2\sqrt{10})(\sqrt{5}+\sqrt{2})^2=(7-2\sqrt{10})(5+2\sqrt{10}+2)$$

   $$=(7-2\sqrt{10})(7+2\sqrt{10})=7^2-(2\sqrt{10})^2=49-40=9.$$

7. $$\left(\sqrt{12-2\sqrt{11}}+\sqrt{12+2\sqrt{11}}\right)^2$$

   $$=(12-2\sqrt{11})+2\sqrt{(12-2\sqrt{11})(12+2\sqrt{11})}+(12+2\sqrt{11})$$

   $$=24+2\sqrt{144-44}=24+2\sqrt{100}=24+20=44.$$

8. $$\left(\sqrt{9+3\sqrt{5}}-\sqrt{9-3\sqrt{5}}\right)^2$$

   $$=(9+3\sqrt{5})-2\sqrt{(9+3\sqrt{5})(9-3\sqrt{5})}+(9-3\sqrt{5})$$

   $$=18-2\sqrt{81-45}=18-2\sqrt{36}=18-12=6.$$