Упр.383 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1. $$0,(2)=0,2222\ldots=\frac{2}{9}$$

   $$2,(21)=2,2121\ldots=2+\frac{21}{99}=2+\frac{7}{33}=\frac{73}{33}$$

   $$15,3(53)=15,35353\ldots=15+\frac{353-3}{990}=15+\frac{35}{99}=\frac{1520}{99}$$

   $$-2,77(3)=-2,77333\ldots=-2-\frac{773-77}{900}=-2-\frac{29}{36}=-\frac{101}{36}$$

2. $$3\sqrt{20}+\sqrt{28}+\sqrt{45}-\sqrt{63}=3\cdot 2\sqrt{5}+2\sqrt{7}+3\sqrt{5}-3\sqrt{7}$$

   $$=6\sqrt{5}+3\sqrt{5}-\sqrt{7}=9\sqrt{5}-\sqrt{7}$$

3. $$\left(2\sqrt{\frac{2}{3}}-8\sqrt{\frac{3}{8}}+3\sqrt{\frac{3}{2}}\right)\cdot 3\sqrt{\frac{3}{2}}$$

   $$=2\cdot 3\sqrt{\frac{2}{3}\cdot\frac{3}{2}}-8\cdot 3\sqrt{\frac{3}{8}\cdot\frac{3}{2}}+3\cdot 3\sqrt{\frac{3}{2}\cdot\frac{3}{2}}$$

   $$=6\sqrt{1}-24\sqrt{\frac{9}{16}}+9\cdot\frac{3}{2}=6-24\cdot\frac{3}{4}+13,5=19,5-18=1,5$$

4. $$\left(6\sqrt{45}-3\sqrt{20}+9\sqrt{80}\right):\left(3\sqrt{5}\right)$$

   $$=\left(6\cdot 3\sqrt{5}-3\cdot 2\sqrt{5}+9\cdot 4\sqrt{5}\right):\left(3\sqrt{5}\right)$$

   $$=\left(18\sqrt{5}-6\sqrt{5}+36\sqrt{5}\right):\left(3\sqrt{5}\right)=48\sqrt{5}:3\sqrt{5}=16$$

5. $$\left(7\sqrt{8}-14\sqrt{18}+0,7\sqrt{12}\right):\left(7\sqrt{2}\right)$$

   $$=\frac{7\sqrt{2}\cdot 2\sqrt{4}}{7\sqrt{2}}-\frac{14\sqrt{2}\cdot 3\sqrt{9}}{7\sqrt{2}}+\frac{0,7\sqrt{4}\cdot 3\sqrt{2}}{7\sqrt{2}}$$

   $$=2-2\cdot 3+0,1\sqrt{6}=2-6+0,1\sqrt{6}=-4+0,1\sqrt{6}$$

6. $$\frac{5}{1+\sqrt{6}}+\frac{6}{3+\sqrt{6}}=\frac{5(1-\sqrt{6})}{1-6}+\frac{6(3-\sqrt{6})}{9-6}$$

   $$=\frac{5(1-\sqrt{6})}{-5}+\frac{6(3-\sqrt{6})}{3}=\sqrt{6}-1+2(3-\sqrt{6})$$

   $$=\sqrt{6}-1+6-2\sqrt{6}=5-\sqrt{6}$$

7. $$\frac{6}{\sqrt{2}-\sqrt{3}}-\frac{4}{\sqrt{2}+\sqrt{3}}=\frac{6(\sqrt{2}+\sqrt{3})-4(\sqrt{2}-\sqrt{3})}{2-3}$$

   $$=\frac{6\sqrt{2}+6\sqrt{3}-4\sqrt{2}+4\sqrt{3}}{-1}=-(2\sqrt{2}+10\sqrt{3})$$

   $$=-2\sqrt{2}-10\sqrt{3}$$

Ответ

1) $$\frac{2}{9},\ \frac{73}{33},\ \frac{1520}{99},\ -\frac{101}{36}$$; 2) $$9\sqrt{5}-\sqrt{7}$$; 3) $$1,5$$; 4) $$16$$; 5) $$-4+0,1\sqrt{6}$$; 6) $$5-\sqrt{6}$$; 7) $$-2\sqrt{2}-10\sqrt{3}$$.