Упр.91 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

Вычислить:

1. 1) $$\left(\frac{1}{16}\right)^{-0{,}75}+10\,000^{0{,}25}-\left(7\frac{19}{32}\right)^{\frac{1}{5}}$$; 2) $$\left(0{,}001\right)^{-\frac{1}{3}}-2^{-2}\cdot64^{\frac{2}{3}}-8^{-1\frac{1}{3}}$$; 3) $$27^{\frac{2}{3}}-\left(-2\right)^{-2}+\left(3\frac{3}{8}\right)^{-\frac{1}{3}}$$; 4) $$\left(-0{,}5\right)^{-4}-625-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}$$.

1. $$\left(\frac{1}{16}\right)^{-0,75}+10000^{0,25}-\left(7\frac{19}{32}\right)^{\frac15}$$
   $$=\left(\frac{1}{16}\right)^{-\frac34}+(10^4)^{\frac14}-\left(\frac{243}{32}\right)^{\frac15}$$
   $$=(2^4)^{\frac34}+10-\left(\frac{3^5}{2^5}\right)^{\frac15}$$
   $$=2^3+10-\frac32=8+10-1,5=16,5.$$

2. $$\left(0,001\right)^{-\frac13}-2^{-2}\cdot 64^{\frac23}-8^{-1\frac13}$$
   $$=\left(\frac{1}{1000}\right)^{-\frac13}-\left(\frac12\right)^2\cdot (4^3)^{\frac23}-8^{-\frac43}$$
   $$=(10^3)^{\frac13}-\frac14\cdot 4^2-(2^3)^{-\frac43}$$
   $$=10-4-\left(\frac12\right)^4=6-\frac{1}{16}=5\frac{15}{16}.$$

3. $$27^{\frac23}-(-2)^{-2}+\left(3\frac38\right)^{-\frac13}$$
   $$=(3^3)^{\frac23}-\left(-\frac12\right)^2+\left(\frac{27}{8}\right)^{-\frac13}$$
   $$=3^2-\frac14+\left(\frac{3^3}{2^3}\right)^{-\frac13}$$
   $$=9-\frac14+\left(\frac32\right)^{-1}$$
   $$=9-\frac14+\frac23=\frac{35}{4}+\frac23=\frac{113}{12}=9\frac{5}{12}.$$

4. $$(-0,5)^{-4}-625-\left(2\frac14\right)^{-1\frac12}$$
   $$=\left(-\frac{5}{10}\right)^{-4}-625-\left(\frac94\right)^{-\frac32}$$
   $$=\left(-\frac{10}{5}\right)^4-625-\left(\frac{3^2}{2^2}\right)^{-\frac32}$$
   $$=(-2)^4-625-\left(\frac32\right)^{-3}$$
   $$=16-625-\frac{8}{27}=-609-\frac{8}{27}=-609\frac{8}{27}.$$

Ответ

1) $$16,5$$; 2) $$5\frac{15}{16}$$; 3) $$9\frac{5}{12}$$; 4) $$-609\frac{8}{27}$$.