Упр.101 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

1. Выполните умножение:
   1) $$(\sqrt{45}+\sqrt{180})\cdot\sqrt{5};$$
   2) $$(6\sqrt{2}-3\sqrt{50}+\sqrt{72})\cdot\sqrt{2};$$
   3) $$(4-\sqrt{6})(2+3\sqrt{6});$$
   4) $$(2\sqrt{7}-\sqrt{2})(\sqrt{7}+6\sqrt{2});$$
   5) $$(\sqrt{19}-\sqrt{13})(\sqrt{19}+\sqrt{13});$$
   6) $$(6\sqrt{m}+8\sqrt{n})(6\sqrt{m}-8\sqrt{n});$$
   7) $$(\sqrt{3}+2)^2;$$
   8) $$(2\sqrt{6}-3\sqrt{7})^2.$$

1. $$\left(\sqrt{45}+\sqrt{180}\right)\cdot\sqrt{5}=\left(\sqrt{9\cdot 5}+\sqrt{36\cdot 5}\right)\cdot\sqrt{5}$$
   $$=\left(3\sqrt{5}+6\sqrt{5}\right)\cdot\sqrt{5}=9\sqrt{5}\cdot\sqrt{5}=9\cdot 5=45.$$
2. $$\left(6\sqrt{2}-3\sqrt{50}+\sqrt{72}\right)\cdot\sqrt{2}=\left(6\sqrt{2}-3\sqrt{25\cdot 2}+\sqrt{36\cdot 2}\right)\cdot\sqrt{2}$$
   $$=\left(6\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\cdot\sqrt{2}=-3\sqrt{2}\cdot\sqrt{2}=-3\cdot 2=-6.$$
3. $$\left(4-\sqrt{6}\right)\left(2+3\sqrt{6}\right)=8+12\sqrt{6}-2\sqrt{6}-3\sqrt{6}\cdot\sqrt{6}$$
   $$=8+10\sqrt{6}-3\cdot 6=8+10\sqrt{6}-18=10\sqrt{6}-10.$$
4. $$\left(2\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+6\sqrt{2}\right)=2\sqrt{7}\cdot\sqrt{7}+12\sqrt{14}-\sqrt{14}-6\sqrt{2}\cdot\sqrt{2}$$
   $$=2\cdot 7+12\sqrt{14}-\sqrt{14}-6\cdot 2=14+11\sqrt{14}-12=2+11\sqrt{14}.$$
5. $$\left(\sqrt{19}-\sqrt{13}\right)\left(\sqrt{19}+\sqrt{13}\right)=19-13=6.$$
6. $$\left(6\sqrt{m}+8\sqrt{n}\right)\left(6\sqrt{m}-8\sqrt{n}\right)=36m-64n.$$
7. $$\left(\sqrt{3}+2\right)^2=3+4\sqrt{3}+4=7+4\sqrt{3}.$$
8. $$\left(2\sqrt{6}-3\sqrt{7}\right)^2=4\cdot 6-2\cdot 2\sqrt{6}\cdot 3\sqrt{7}+9\cdot 7$$
   $$=24-12\sqrt{42}+63=87-12\sqrt{42}.$$

Ответ: 1) $$45$$; 2) $$-6$$; 3) $$10\sqrt{6}-10$$; 4) $$2+11\sqrt{14}$$; 5) $$6$$; 6) $$36m-64n$$; 7) $$7+4\sqrt{3}$$; 8) $$87-12\sqrt{42}$$.