Упр.22.3 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) 2^(1/4)·8^(1/4); 2) 0,054^(1/3)·4^(1/3);
3) (4^(1/5))/128^(1/5); 4) ((2^30·7^12)^(1/8))/(2^6·7^4)^(1/8);
5) (6v3+10)^(1/3)·(6v3-10)^(1/3); 6) 3^(1/4)·3^(1/3)·27^(1/4)·(-9)^(1/3).

1. $$2^{\frac14}\cdot 8^{\frac14}=(2\cdot 8)^{\frac14}=16^{\frac14}=2.$$
2. $$0{,}054^{\frac13}\cdot 4^{\frac13}=(0{,}054\cdot 4)^{\frac13}=0{,}216^{\frac13}=0{,}6.$$
3. $$\frac{4^{\frac15}}{128^{\frac15}}=\left(\frac{4}{128}\right)^{\frac15}=\left(\frac{1}{32}\right)^{\frac15}=\left(\frac{1}{2^5}\right)^{\frac15}=\frac12.$$
4. $$\frac{(2^{30}\cdot 7^{12})^{\frac18}}{(2^6\cdot 7^4)^{\frac18}}=\left(\frac{2^{30}\cdot 7^{12}}{2^6\cdot 7^4}\right)^{\frac18}=\left(2^{24}\cdot 7^8\right)^{\frac18}=2^3\cdot 7=56.$$
5. $$\left(6\sqrt3+10\right)^{\frac13}\cdot \left(6\sqrt3-10\right)^{\frac13}=\sqrt[3]{(6\sqrt3+10)(6\sqrt3-10)}$$
   $$=\sqrt[3]{(6\sqrt3)^2-10^2}=\sqrt[3]{108-100}=\sqrt[3]{8}=2.$$
6. $$3^{\frac14}\cdot 3^{\frac13}\cdot 27^{\frac14}\cdot (-9)^{\frac13} =3^{\frac14}\cdot 3^{\frac13}\cdot (3^3)^{\frac14}\cdot \left((-3)^2\right)^{\frac13}$$
   $$=3^{\frac14}\cdot 3^{\frac13}\cdot 3^{\frac34}\cdot (-3^{\frac23}) =-3^{\frac14+\frac13+\frac34+\frac23}=-3^3=-9.$$

Ответ

1) 2; 2) 0,6; 3) 0,5; 4) 56; 5) 2; 6) −9.