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

1) v(v82-1)•v(v82+1);
2) v(v58+v33) •v(v58-v33) ;
3) (v2+3)^2-(5-2v2)^2;
4) (4+3v6)^2+(4-3v6)^2;
5) (16+6v7) (3-v7)^2;
6) (7-2v10) (v5+v2)^2;
7) (v(12-2v11) +v(12+2v11) )^2;
8) (v(9+3v5) -v(9-3v5) )^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. $$ (\sqrt{2}+3)^2-(5-2\sqrt{2})^2 $$
   $$ =\left(2+6\sqrt{2}+9\right)-\left(25-20\sqrt{2}+8\right) $$
   $$ =11+6\sqrt{2}-33+20\sqrt{2} =26\sqrt{2}-22. $$

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. $$ \left(\sqrt{9+3\sqrt{5}}-\sqrt{9-3\sqrt{5}}\right)^2 $$
   $$ =(9+3\sqrt{5})-2\sqrt{(9+3\sqrt{5})(9-3\sqrt{5})}+(9-3\sqrt{5}) $$
   $$ =18-2\sqrt{81-45}=18-2\sqrt{36}=18-12=6. $$

Ответ

1) $$9$$; 2) $$5$$; 3) $$26\sqrt{2}-22$$; 4) $$140$$; 5) $$4$$; 6) $$9$$; 7) $$44$$; 8) $$6$$.