Question

Умоляю помогите даю 50 баллов! Срочно... ​

Answer 1

Ответ:

Объяснение:

Формулы :

$1)\ \sf\dfrac{1}{a} =a^{-1}} \\\\\\ 2) \ \sf (-a)^{-2}=\dfrac{1}{(-a)\cdot (-a)} }=\dfrac{1}{a^2}$

Решение:

$\tt \displaystyle 2)\ 6^{-3}=\frac{1}{216} \\\\\\ 3)\ (-5)^{-4}=\frac{1}{625}$      $\tt 5)\ \bigg( -\dfrac{1}{9} \bigg)^{-1}=-9 \\\\\\ 6) \bigg(\dfrac{5}{7} \bigg)^{-3}=\bigg(\dfrac{7}{5} \bigg)^3=\dfrac{343}{125} =2\dfrac{93}{125}$      

$\displaystyle \tt 8) \ \bigg(1\frac{3}{4} \bigg)^{-1}=\frac{4}{7} \\\\\\ 9) \ 0,7^{-2}=\bigg(\frac{7}{10} \bigg)^{-2} =\frac{100}{49} =2\frac{2}{49}$

№ 44

$\displaystyle \sf 1) \ 5^{-2}+10^{-2}=\frac{1}{25} +\frac{1}{100} =\frac{5}{100} =\frac{1}{20} \\\\\\ 2) \bigg(\frac{6}{7} \bigg)^{-1}+6^{-2}-(-3,5)^2=\\\\ 3,5=\frac{21}{6} \\\\\\ \frac{7}{6}^{/6} +\frac{1}{36} -\bigg(-\frac{21}{6}\bigg)^2 =\frac{43-441}{36} =\frac{398}{36} =11\frac{1}{18} \\\\\\ 3) \ \bigg(\frac{9}{4} \bigg)^{-2}\cdot 2^{-5}=\frac{16}{81} \cdot \frac{1}{32} =\frac{1}{162}$