Упр.559 ГДЗ Мерзляк Полонский 8 класс (Алгебра)
- Упростите выражение:
1) $$8\sqrt{2}-\sqrt{32}$$;
2) $$6\sqrt{3}-\sqrt{27}$$;
3) $$\sqrt{96}-3\sqrt{6}$$;
4) $$2\sqrt{500}-8\sqrt{5}$$;
5) $$5\sqrt{7}-\sqrt{700}-0{,}5\sqrt{28}$$;
6) $$2\sqrt{20}-\frac{1}{3}\sqrt{45}-0{,}6\sqrt{125}$$. - Докажите равенство:
1) $$\frac{1}{5-2\sqrt{6}}+\frac{1}{5+2\sqrt{6}}=10$$;
2) $$\frac{2}{3\sqrt{2}+4}+\frac{2}{3\sqrt{2}-4}=-8$$;
3) $$\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}=4\sqrt{2}$$.
$$8\sqrt{2}-\sqrt{32}=8\sqrt{2}-\sqrt{16\cdot 2}=8\sqrt{2}-4\sqrt{2}=4\sqrt{2}.$$
$$6\sqrt{3}-\sqrt{27}=6\sqrt{3}-\sqrt{9\cdot 3}=6\sqrt{3}-3\sqrt{3}=3\sqrt{3}.$$
$$\sqrt{96}-3\sqrt{6}=\sqrt{16\cdot 6}-3\sqrt{6}=4\sqrt{6}-3\sqrt{6}=\sqrt{6}.$$
$$2\sqrt{500}-8\sqrt{5}=2\sqrt{100\cdot 5}-8\sqrt{5}=2\cdot 10\sqrt{5}-8\sqrt{5}=12\sqrt{5}.$$
$$5\sqrt{7}-\sqrt{700}-0{,}5\sqrt{28}=5\sqrt{7}-\sqrt{7\cdot 100}-0{,}5\sqrt{7\cdot 4}$$
$$=5\sqrt{7}-10\sqrt{7}-0{,}5\cdot 2\sqrt{7}=-5\sqrt{7}-\sqrt{7}=-6\sqrt{7}.$$$$2\sqrt{20}-\frac{1}{3}\sqrt{45}-0{,}6\sqrt{125}=2\sqrt{4\cdot 5}-\frac{1}{3}\sqrt{9\cdot 5}-0{,}6\sqrt{25\cdot 5}$$
$$=2\cdot 2\sqrt{5}-\frac{1}{3}\cdot 3\sqrt{5}-0{,}6\cdot 5\sqrt{5}=4\sqrt{5}-\sqrt{5}-3\sqrt{5}=0.$$
$$\frac{1}{5-2\sqrt{6}}+\frac{1}{5+2\sqrt{6}}=\frac{(5+2\sqrt{6})+(5-2\sqrt{6})}{(5-2\sqrt{6})(5+2\sqrt{6})}$$
$$=\frac{10}{25-24}=\frac{10}{1}=10.$$$$\frac{2}{3\sqrt{2}+4}+\frac{2}{3\sqrt{2}-4}=\frac{2(3\sqrt{2}-4)+2(3\sqrt{2}+4)}{(3\sqrt{2}+4)(3\sqrt{2}-4)}$$
$$=\frac{6\sqrt{2}-8+6\sqrt{2}+8}{18-16}=\frac{12\sqrt{2}}{2}=6\sqrt{2}.$$$$\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)^2+(\sqrt{2}-1)^2}{(\sqrt{2}-1)(\sqrt{2}+1)}$$
$$=\frac{(2+2\sqrt{2}+1)+(2-2\sqrt{2}+1)}{2-1}=\frac{6}{1}=6.$$
Ответ
1) $$4\sqrt{2}$$; 2) $$3\sqrt{3}$$; 3) $$\sqrt{6}$$; 4) $$12\sqrt{5}$$; 5) $$-6\sqrt{7}$$; 6) $$0$$; 7) $$10$$; 8) $$6\sqrt{2}$$; 9) $$6$$.
