Упр.167 ГДЗ Никольский Потапов 8 класс (Алгебра)

1. $$2\sqrt{2}+3\sqrt{2}=5\sqrt{2}.$$
2. $$2\sqrt{8}-3\sqrt{2}=2\cdot 2\sqrt{2}-3\sqrt{2}=\sqrt{2}.$$
3. $$\sqrt{a}-5\sqrt{a}=-4\sqrt{a}.$$
4. $$a\sqrt{x}-3\sqrt{x}=(a-3)\sqrt{x}.$$
5. $$2\sqrt{a}+3\sqrt{a}-\sqrt{4a}=2\sqrt{a}+3\sqrt{a}-2\sqrt{a}=3\sqrt{a}.$$
6. $$\sqrt{2}+3\sqrt{32}+\frac{1}{2}\sqrt{128}-6\sqrt{18} =\sqrt{2}+3\cdot 4\sqrt{2}+\frac{1}{2}\cdot 8\sqrt{2}-6\cdot 3\sqrt{2} =-\sqrt{2}.$$
7. $$(8+3\sqrt{5})(2-\sqrt{5}) =16-8\sqrt{5}+6\sqrt{5}-15 =1-2\sqrt{5}.$$
8. $$(3\sqrt{8}+\sqrt{18}+\sqrt{50}-2\sqrt{72})\cdot \sqrt{2} =(6\sqrt{2}+3\sqrt{2}+5\sqrt{2}-12\sqrt{2})\cdot \sqrt{2} =2\sqrt{2}\cdot \sqrt{2}=4.$$
9. $$(3-\sqrt{2})(2+3\sqrt{2}) =6+9\sqrt{2}-2\sqrt{2}-6 =7\sqrt{2}.$$
10. $$(7\sqrt{2}-5\sqrt{6}-3\sqrt{8}+4\sqrt{20})\cdot 3\sqrt{2} =(7\sqrt{2}-5\sqrt{6}-6\sqrt{2}+8\sqrt{5})\cdot 3\sqrt{2}$$
    $$=(\sqrt{2}-5\sqrt{6}+8\sqrt{5})\cdot 3\sqrt{2} =6-30\sqrt{3}+24\sqrt{10}.$$
11. $$(2\sqrt{6}+5\sqrt{3}-7\sqrt{2})(\sqrt{6}-2\sqrt{3}+4\sqrt{2}) =-74+3\sqrt{2}+2\sqrt{3}+34\sqrt{6}.$$
12. $$(\sqrt{12}-1)(\sqrt{12}+1)=(\sqrt{12})^2-1^2=12-1=11.$$
13. $$(7-\sqrt{3})(\sqrt{3}+7)=7^2-(\sqrt{3})^2=49-3=46.$$
14. $$(\sqrt{20}-3)(3+2\sqrt{5})=(\sqrt{20}-3)(\sqrt{20}+3)=(\sqrt{20})^2-3^2=20-9=11.$$

Ответ

а) $$5\sqrt{2}$$; б) $$\sqrt{2}$$; в) $$-4\sqrt{a}$$; г) $$(a-3)\sqrt{x}$$; д) $$3\sqrt{a}$$; е) $$-\sqrt{2}$$; ж) $$1-2\sqrt{5}$$; з) $$4$$; и) $$7\sqrt{2}$$; к) $$6-30\sqrt{3}+24\sqrt{10}$$; л) $$-74+3\sqrt{2}+2\sqrt{3}+34\sqrt{6}$$; м) $$11$$; н) $$46$$; о) $$11$$.