Упр.47 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

а) v(v5-2,5)^2-((1,5-v5)^3)^(1/3)-1;
б) (5v3+v50)(5-v24)/(v75-5v2);
в) (v(v2-1,5)^2-((1-v2)^3)^(1/3))^2+0,75;
г) (2v6-v20)/(2v5+v24)·(11+2v30).

1. а)

   $$\sqrt{(\sqrt5-2{,}5)^2}-\sqrt[3]{(1{,}5-\sqrt5)^3}-1$$
   $$=(2{,}5-\sqrt5)-(1{,}5-\sqrt5)-1$$
   $$=0.$$

2. б)

   $$\frac{(5\sqrt3+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt2} =\frac{5(\sqrt3+\sqrt2)(5-2\sqrt6)}{5(\sqrt3-\sqrt2)}$$
   $$=\frac{5(3+2\sqrt6+2)(5-2\sqrt6)}{5(3-2)}$$
   $$=\frac{5(25-4\cdot 6)}{5}=1.$$

3. в)

   $$\left(\sqrt{(\sqrt2-1{,}5)^2}-\sqrt[3]{(1-\sqrt2)^3}\right)^2+0{,}75$$
   $$=\left((1{,}5-\sqrt2)-(1-\sqrt2)\right)^2+0{,}75$$
   $$=(0{,}5)^2+0{,}75=0{,}25+0{,}75=1.$$

4. г)

   $$\frac{2\sqrt6-\sqrt{20}}{2\sqrt5+\sqrt{24}}\cdot(11+2\sqrt{30}) =\frac{2\sqrt6-2\sqrt5}{2\sqrt5+2\sqrt6}\cdot(11+2\sqrt{30})$$
   $$=\frac{(\sqrt6-\sqrt5)^2}{6-5}\cdot(11+2\sqrt{30})$$
   $$=(6-2\sqrt{30}+5)(11+2\sqrt{30})$$
   $$=(11-2\sqrt{30})(11+2\sqrt{30})$$
   $$=121-4\cdot 30=1.$$

Ответ

а) $$0$$; б) $$1$$; в) $$1$$; г) $$1$$.