Упр.22.9 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) 6v3 + v75 — v243; г) 7v5 — v80 + v180;
б) 0,5v125 + 2v320 — 6v20; д) 2v75 + 13v3 — 1/3 v27;
в) 2/3 v243 — 3/4 v192 + 1/6 v75; е) 3/5 v 275 + 4/7 v539 — 5/6 v44.

1. а)

   $$6\sqrt{3}+\sqrt{75}-\sqrt{243}=6\sqrt{3}+\sqrt{25\cdot 3}-\sqrt{81\cdot 3}$$
   $$=6\sqrt{3}+5\sqrt{3}-9\sqrt{3}=2\sqrt{3}.$$

2. б)

   $$0{,}5\sqrt{125}+2\sqrt{320}-6\sqrt{20}=0{,}5\sqrt{25\cdot 5}+2\sqrt{64\cdot 5}-6\sqrt{4\cdot 5}$$
   $$=0{,}5\cdot 5\sqrt{5}+2\cdot 8\sqrt{5}-6\cdot 2\sqrt{5}$$
   $$=2{,}5\sqrt{5}+16\sqrt{5}-12\sqrt{5}=6{,}5\sqrt{5}.$$

3. в)

   $$\frac{2}{3}\sqrt{243}-\frac{3}{4}\sqrt{192}+\frac{1}{6}\sqrt{75} =\frac{2}{3}\sqrt{81\cdot 3}-\frac{3}{4}\sqrt{64\cdot 3}+\frac{1}{6}\sqrt{25\cdot 3}$$
   $$=\frac{2}{3}\cdot 9\sqrt{3}-\frac{3}{4}\cdot 8\sqrt{3}+\frac{1}{6}\cdot 5\sqrt{3}$$
   $$=6\sqrt{3}-6\sqrt{3}+\frac{5}{6}\sqrt{3}=\frac{5}{6}\sqrt{3}.$$

4. г)

   $$7\sqrt{5}-\sqrt{80}+\sqrt{180}=7\sqrt{5}-\sqrt{16\cdot 5}+\sqrt{36\cdot 5}$$
   $$=7\sqrt{5}-4\sqrt{5}+6\sqrt{5}=9\sqrt{5}.$$

5. д)

   $$2\sqrt{75}+13\sqrt{3}-\frac{1}{3}\sqrt{27}=2\sqrt{25\cdot 3}+13\sqrt{3}-\frac{1}{3}\sqrt{9\cdot 3}$$
   $$=2\cdot 5\sqrt{3}+13\sqrt{3}-\frac{1}{3}\cdot 3\sqrt{3}$$
   $$=10\sqrt{3}+13\sqrt{3}-\sqrt{3}=22\sqrt{3}.$$

6. е)

   $$\frac{3}{5}\sqrt{275}+\frac{4}{7}\sqrt{539}-\frac{5}{6}\sqrt{44} =\frac{3}{5}\sqrt{25\cdot 11}+\frac{4}{7}\sqrt{49\cdot 11}-\frac{5}{6}\sqrt{4\cdot 11}$$
   $$=\frac{3}{5}\cdot 5\sqrt{11}+\frac{4}{7}\cdot 7\sqrt{11}-\frac{5}{6}\cdot 2\sqrt{11}$$
   $$=3\sqrt{11}+4\sqrt{11}-\frac{5}{3}\sqrt{11}=7\sqrt{11}-\frac{5}{3}\sqrt{11}=\frac{16}{3}\sqrt{11}.$$

Ответ

а) $$2\sqrt{3}$$; б) $$6{,}5\sqrt{5}$$; в) $$\frac{5}{6}\sqrt{3}$$; г) $$9\sqrt{5}$$; д) $$22\sqrt{3}$$; е) $$\frac{16}{3}\sqrt{11}$$.