Упр.76 ГДЗ Алимов 10-11 класс (Алгебра)

Задача

1)(1/16)^-0,75 + 810 000^0,25 — (7*19/32)1/5;
2) 27^2/3 — (-2)^-2 + (3*3/8)^-1/3;
3)(0,001)^-1/3 — 2^-2 *64^2/3 — 8^(-1*1/3);
4) (-0,5)^-4 — 625^0,25 — (2*1/4)^-1*1/2.

Подробный ответ

$$\left(\frac{1}{16}\right)^{-0,75}+810000^{0,25}-\left(7\frac{19}{32}\right)^{\frac15}$$
$$\left(\frac{1}{16}\right)^{-0,75}=\left(2^{-4}\right)^{-\frac34}=2^3=8,$$
$$810000^{0,25}=(81\cdot 10000)^{\frac14}=3^2\cdot 10=30,$$
$$\left(7\frac{19}{32}\right)^{\frac15}=\left(\frac{243}{32}\right)^{\frac15}=\frac{3}{2}.$$
$$8+30-\frac32=38-\frac32=36,5.$$
$$27^{\frac23}-(-2)^{-2}+\left(3\frac38\right)^{-\frac13}$$
$$27^{\frac23}=(3^3)^{\frac23}=3^2=9,$$
$$(-2)^{-2}=\frac{1}{(-2)^2}=\frac14,$$
$$\left(3\frac38\right)^{-\frac13}=\left(\frac{27}{8}\right)^{-\frac13}=\left(\frac{8}{27}\right)^{\frac13}=\frac23.$$
$$9-\frac14+\frac23=\frac{108-3+8}{12}=\frac{113}{12}=9\frac{5}{12}.$$
$$(0,001)^{-\frac13}-2^{-2}\cdot 64^{\frac23}-8^{-1\frac13}$$
$$(0,001)^{-\frac13}=\left(\frac{1}{1000}\right)^{-\frac13}=10,$$
$$2^{-2}\cdot 64^{\frac23}=\frac14\cdot (2^6)^{\frac23}=\frac14\cdot 2^4=4,$$
$$8^{-1\frac13}=8^{-\frac43}=\left(2^3\right)^{-\frac43}=2^{-4}=\frac{1}{16}.$$
$$10-4-\frac{1}{16}=6-\frac{1}{16}=5\frac{15}{16}.$$
$$(-0,5)^{-4}-625^{0,25}-\left(2\frac14\right)^{-1\frac12}$$
$$(-0,5)^{-4}=\left(-\frac12\right)^{-4}=16,$$
$$625^{0,25}=625^{\frac14}=5,$$
$$\left(2\frac14\right)^{-1\frac12}=\left(\frac94\right)^{-\frac32}=\left(\frac49\right)^{\frac32}=\left(\frac23\right)^3=\frac{8}{27}.$$
$$16-5-\frac{8}{27}=11-\frac{8}{27}=10\frac{19}{27}.$$