Упр.86 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) (0,175)^0+(0,36)^(-2)-1^(4/3);
2) 1^(-0,43)-(0,008)^(-1/3)+(15,1)^0;
3) (4/5)^(-2)-(1/27)^(1/3)+4·379^(0);
4) (0,125)^(-1/3)+(3/4)^2-(1,85)^0.

1. $$ (0{,}175)^0+(0{,}36)^{-2}-1^{4/3}=1+\left(\frac{36}{100}\right)^{-2}-1 $$
   $$ =\left(\frac{9}{25}\right)^{-2}=\left(\frac{25}{9}\right)^2=\frac{625}{81}=7\frac{58}{81}. $$

2. $$ 1^{-0{,}43}-(0{,}008)^{-1/3}+(15{,}1)^0=1-\left(\frac{8}{1000}\right)^{-1/3}+1 $$
   $$ =2-\left(\frac{2}{10}\right)^{-1}=2-5=-3. $$

3. $$ \left(\frac45\right)^{-2}-\left(\frac1{27}\right)^{1/3}+4\cdot 379^0=\left(\frac54\right)^2-\frac13+4\cdot 1 $$
   $$ =\frac{25}{16}-\frac13+4=\frac{75-16+192}{48}=\frac{251}{48}=5\frac{11}{48}. $$

4. $$ (0{,}125)^{-1/3}+\left(\frac34\right)^2-(1{,}85)^0=\left(\frac{125}{1000}\right)^{-1/3}+\frac{9}{16}-1 $$
   $$ =\left(\frac{5}{10}\right)^{-1}+\frac{9}{16}-1=2+\frac{9}{16}-1=\frac{25}{16}=1\frac{9}{16}. $$

Ответ

1) $$7\frac{58}{81}$$; 2) $$-3$$; 3) $$5\frac{11}{48}$$; 4) $$1\frac{9}{16}$$.