Упр.14 Повторение ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

Используем свойство корня: $$\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}.$$

1. $$10\sqrt{\frac{2}{5}}-0{,}5\sqrt{160}+\frac{1}{9}\sqrt{81}$$

   $$10\sqrt{\frac{2}{5}}=10\sqrt{\frac{2\cdot 5}{5\cdot 5}}=10\cdot\frac{\sqrt{10}}{5}=2\sqrt{10},$$

   $$0{,}5\sqrt{160}=0{,}5\sqrt{16\cdot 10}=0{,}5\cdot 4\sqrt{10}=2\sqrt{10},$$

   $$\frac{1}{9}\sqrt{81}=\frac{1}{9}\cdot 9=1.$$

   Тогда

   $$2\sqrt{10}-2\sqrt{10}+1=1.$$

2. $$4\sqrt{3\frac{1}{2}}-0{,}5\sqrt{56}-\sqrt{4\frac{1}{4}}$$

   $$3\frac{1}{2}=\frac{7}{2}, \qquad 4\frac{1}{4}=\frac{17}{4},$$

   $$4\sqrt{\frac{7}{2}}=4\cdot\frac{\sqrt{14}}{2}=2\sqrt{14},$$

   $$0{,}5\sqrt{56}=0{,}5\sqrt{4\cdot 14}=0{,}5\cdot 2\sqrt{14}=\sqrt{14},$$

   $$\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}.$$

   Следовательно,

   $$2\sqrt{14}-\sqrt{14}-\frac{\sqrt{17}}{2}=\sqrt{14}-\frac{\sqrt{17}}{2}.$$

3. $$16\sqrt{\frac{3}{5}}-9\sqrt{60}+2\sqrt{3\frac{3}{4}}$$

   $$16\sqrt{\frac{3}{5}}=16\sqrt{\frac{15}{25}}=16\cdot\frac{\sqrt{15}}{5}=\frac{16\sqrt{15}}{5},$$

   $$9\sqrt{60}=9\sqrt{4\cdot 15}=18\sqrt{15},$$

   $$2\sqrt{3\frac{3}{4}}=2\sqrt{\frac{15}{4}}=2\cdot\frac{\sqrt{15}}{2}=\sqrt{15}.$$

   Тогда

   $$\frac{16\sqrt{15}}{5}-18\sqrt{15}+\sqrt{15}=\frac{16\sqrt{15}}{5}-17\sqrt{15}=-\frac{69\sqrt{15}}{5}.$$

4. $$3\sqrt{2\frac{1}{3}}-\sqrt{84}-\sqrt{5\frac{1}{4}}$$

   $$2\frac{1}{3}=\frac{7}{3}, \qquad 5\frac{1}{4}=\frac{21}{4},$$

   $$3\sqrt{\frac{7}{3}}=3\cdot\frac{\sqrt{21}}{3}=\sqrt{21},$$

   $$\sqrt{84}=\sqrt{4\cdot 21}=2\sqrt{21},$$

   $$\sqrt{\frac{21}{4}}=\frac{\sqrt{21}}{2}.$$

   Следовательно,

   $$\sqrt{21}-2\sqrt{21}-\frac{\sqrt{21}}{2}=-\frac{3\sqrt{21}}{2}.$$

Ответ: 1) $$1$$; 2) $$\sqrt{14}-\frac{\sqrt{17}}{2}$$; 3) $$-\frac{69\sqrt{15}}{5}$$; 4) $$-\frac{3\sqrt{21}}{2}$$.