Question

помогитее прошу, мне рили это нужно ​

Answer 1

$\displaystyle1) { \bigg( - \frac{1}{3} \bigg)}^{ - 2} = \frac{ {3}^{2} }{ {1}^{2} } = \boxed{ \bf9}; \: \: \: 2) { \bigg( \frac{4}{3} \bigg) }^{ - 1} = \boxed{ \bf \frac{3}{4} };$

$3) {0.1}^{ - 2} = \frac{ {1}^{2} }{ {0.1}^{2} } = \boxed{ \bf \frac{1}{0.01} }; \: \: \: 4)2 { \bigg( \frac{1}{3} \bigg) }^{ - 4} = 2 \cdot \frac{ {3}^{4} }{ {1}^{4} } = \boxed{ \bf162};$

$5) {(0.001)}^{ - 3} = \boxed{ \bf\frac{ 1}{ {0.001}^{3} } }; \: \: 6) { \bigg( - 1 \frac{3}{4} \bigg) }^{2} = { \bigg( - \frac{7}{4} \bigg) }^{2} = \frac{49}{16} = \boxed{ \bf3 \frac{1}{16} }; \: \: 7) {(0.25)}^{ - 2} = \frac{ {1}^{2} }{ {0.25}^{2} } = \frac{1}{0.0625} = \boxed{ \bf16};$

$8) { \bigg( - \frac{1}{5} \bigg)}^{ - 5} = { - 5}^{5} = \boxed{ \bf3125};$

$9) { \bigg( \frac{1}{3} + \frac{3}{2} \bigg)}^{2} = { \bigg( \frac{11}{6} \bigg) }^{2} = \frac{121}{36} = \boxed{ \bf3 \frac{13}{36} }.$