Упр.9.27 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) (5(4^(1/3))+0,5(108^(1/3))-500^(1/3))2^(1/3);
2) (2(2^(1/3))-2(5^(1/3))+100^(1/3))(10^(1/3)+4^(1/3));
3) ((9^(1/3)-6(72^(1/3))+2((1125^(1/3)))/9^(1/3);
4) ((2^(1/4)+8^(1/4))^2/(4+3v2).

1. $$\left(5\sqrt[3]{4}+0{,}5\sqrt[3]{108}-\sqrt[3]{500}\right)\sqrt[3]{2}$$
   $$=5\sqrt[3]{8}+0{,}5\sqrt[3]{216}-\sqrt[3]{1000}$$
   $$=5\cdot 2+0{,}5\cdot 6-10=10+3-10=3.$$

2. $$\left(2\sqrt[3]{2}-2\sqrt[3]{5}+\sqrt[3]{100}\right)\left(\sqrt[3]{10}+\sqrt[3]{4}\right)$$
   $$=2\sqrt[3]{20}+2\sqrt[3]{8}-2\sqrt[3]{50}-2\sqrt[3]{20}+\sqrt[3]{1000}+\sqrt[3]{400}$$
   $$=2\sqrt[3]{8}-2\sqrt[3]{50}+10+\sqrt[3]{8\cdot 50}$$
   $$=4-2\sqrt[3]{50}+10+2\sqrt[3]{50}=14.$$

3. $$\frac{\sqrt[3]{9}-6\sqrt[3]{72}+2\sqrt[3]{1125}}{\sqrt[3]{9}}$$
   $$=\sqrt[3]{\frac{9}{9}}-6\sqrt[3]{\frac{72}{9}}+2\sqrt[3]{\frac{1125}{9}}$$
   $$=\sqrt[3]{1}-6\sqrt[3]{8}+2\sqrt[3]{125}$$
   $$=1-6\cdot 2+2\cdot 5=1-12+10=-1.$$

4. $$\frac{\left(\sqrt[4]{2}+\sqrt[4]{8}\right)^2}{4+3\sqrt{2}}$$
   $$=\frac{\sqrt{2}+2\sqrt[4]{16}+\sqrt{8}}{4+3\sqrt{2}}$$
   $$=\frac{\sqrt{2}+2\cdot 2+\sqrt{8}}{4+3\sqrt{2}}$$
   $$=\frac{\sqrt{2}+4+2\sqrt{2}}{4+3\sqrt{2}}=\frac{4+3\sqrt{2}}{4+3\sqrt{2}}=1.$$

Ответ

1) $$3$$; 2) $$14$$; 3) $$-1$$; 4) $$1$$.