Question

ПЖ СРОЧНО 50 БАЛЛОВ !!!!!!!!!!!!!!

Answer 1

Объяснение:

Вспоминаем про то, что

${x}^{ - n} = \frac{1}{ {x}^{n} }$

1)а)

${3}^{ - 2} = \frac{1}{ {3}^{2} } = \frac{1}{9}$

б)

$( - 5) ^{ -2 } = \frac{1}{( - 5)^{2} } = \frac{1}{25}$

в)

${( - 6)}^{ - 1} = \frac{1}{ ( - 6)} = - \frac{1}{6}$

г)

$- 4 \times {( - 2)}^{ - 3} = - 4 \times \frac{1}{ { ( - 2)}^{3} } = \frac{ - 4}{ - 8} = \frac{1}{2}$

2)а)

2)а)

$( \frac{1}{3} )^{ - 1} = \frac{1}{ \frac{1}{3} } = 3$

б)

$( \frac{2}{5} )^{ - 2} = \frac{1}{ {( \frac{2}{5} })^{2} } = \frac{1}{ \frac{4}{25} } = \frac{25}{4} = 6.25$

в)

${( - 0.1)}^{ - 4} = \frac{1}{ {( - 0.1)}^{4} } = \frac{1}{0.0001} = 10000$

г)

${(2 \frac{1}{3}) }^{ - 3} = { (\frac{7}{3} )}^{ - 3} = \frac{1}{ {( \frac{7}{3}) }^{3} } = \frac{1}{ \frac{343}{27} } = \frac{27}{343}$

3)а)

${8}^{ - 1} + {6}^{ - 2} = \frac{1}{8} + \frac{1}{ {6}^{2} } = \frac{1}{8} + \frac{1}{36} = \frac{9 + 2}{72} = \frac{11}{72}$

б)

${2.4}^{ - 1} + {5}^{0} = \frac{1}{2.4} + 1 = \frac{3.4}{2.4} = \frac{ \frac{34}{10} }{ \frac{24}{10} } = \frac{34}{24} = \frac{17}{12} = 1 \frac{5}{12}$

в)

$1453 - {0.1}^{ - 3} = 1453 - \frac{1}{0.1^{3} } = 1453 - \frac{1}{0.001} = 1453 - 1000 = 453$

г)

$18 - {( \frac{1}{5} )}^{ - 2} = 18 - \frac{1}{( \frac{ 1}{5})^{2} } = 18 - \frac{1}{ \frac{1}{25} } = 18 - 25 = - 7$

Answer 2

Ответ:

Пользуемся свойствами степеней с отрицательными показателями:

$\boldsymbol{a^{-1}=\dfrac{1}{a}\ \ ,\ \ \ \ a^{-k}=\dfrac{1}{a^{k}}\ \ ,\ \ \ \ \dfrac{1}{a^{-1}}=a\ \ ,\ \ \ \ \dfrac{1}{a^{-k}}=a^{k}}$   .

$\displaystyle 1)\ \ 3^{-2}=\frac{1}{3^2}=\frac{1}{9}\ \ ,\ \ \ (-5)^{-2}=\frac{1}{(-5)^2}=\frac{1}{25}\ \ ,\ \ \ (-6)^{-1}=\frac{1}{(-6)^1}=-\frac{1}{6}\ ,\\\\\\-4\cdot (-2)^{-3}=-4\cdot\frac{1}{(-2)^3}=-4\cdot \frac{1}{-8}=\frac{4}{8}=\frac{1}{2}$

$\displaystyle 2)\ \ \Big(\frac{1}{3}\Big)^{-1}=3\ \ \ ,\ \ \ \ \ \ \Big(\frac{2}{5}\Big)^{-2}=\Big(\frac{5}{2}\Big)^2=\frac{25}{4}=6,25\ \ ,\\\\\\(-0,1)^{-4}=\Big(-\frac{1}{10}\Big)^{-4}=(-10)^4=10^4=10\, 000\\\\\\\Big(2\frac{1}{3}\Big)^{-3}=\Big(\frac{7}{3}\Big)^{-3}=\Big(\frac{3}{7}\Big)^3=\frac{27}{343}$

$\displaystyle 3)\ \ 8^{-1}+6^{-2}=\frac{1}{8}+\frac{1}{6^2}=\frac{1}{8}+\frac{1}{36}=\frac{9+2}{72}=\frac{11}{72}\\\\\\2,4^{-1}+5^0=\frac{1}{2,4}+1=\frac{10}{24}+1=\frac{5}{12}+1=\frac{5+12}{12}=\frac{17}{12}=1\frac{5}{12}\\\\\\1453-0,1^{-3}=1453-\frac{1}{0,1^3}=1453-\frac{1}{(1/10)^3}=1453-1000=453\\\\\\18-\Big(\frac{1}{5}\Big)^{-2}=18-5^2=18-25=-7$