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

1. Вычислите:
   а) $$\sqrt{(\sqrt{5}-2{,}5)^2}-\sqrt[3]{(1{,}5-\sqrt{5})^3}-1$$;
   б) $$\frac{(5\sqrt{3}+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt{2}}$$;
   в) $$\left(\sqrt{(\sqrt{2}-1{,}5)^2}-\sqrt[3]{(1-\sqrt{2})^3}\right)^2+0{,}75$$;
   г) $$\frac{2\sqrt{6}-\sqrt{20}}{2\sqrt{5}+\sqrt{24}}\cdot(11+2\sqrt{30})$$.

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

$$=(2{,}5-\sqrt5)-(1{,}5-\sqrt5)-1=0.$$

б) $$\frac{(5\sqrt3+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt2}$$

$$=\frac{5(\sqrt3+\sqrt2)(5-2\sqrt6)}{5(\sqrt3-\sqrt2)}$$

$$=\frac{5(3+2\sqrt6+2)(5-2\sqrt6)}{5(3-2)}$$

$$=\frac{5(25-4\cdot 6)}{5}=1.$$

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

$$=\left((1{,}5-\sqrt2)-(1-\sqrt2)\right)^2+0{,}75$$

$$=(0{,}5)^2+0{,}75=0{,}25+0{,}75=1.$$

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) $$1$$.