Question

1.3 Выполните возведение в степень​

Answer 1

Объяснение:

1).

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

2).

$( - \frac{1}{2})^{5} = ( - \frac{1}{2}) \times ( - \frac{1}{2}) \times ( - \frac{1}{2}) \times ( - \frac{1}{2}) \times ( - \frac{1}{2}) = - \frac{1}{32}$

3).

$( - \frac{2}{3})^{2} = ( - \frac{2}{3}) \times ( - \frac{2}{3}) = \frac{4}{9}$

4).

$0.3 ^{3} = 0.3 \times 0.3 \times 0.3 = 0.027$

5).

${0.1}^{3} = 0.1 \times 0.1 \times 0.1 = 0.001$

6).

$(1 \frac{1}{2}) ^{4} = (\frac{3}{2}) ^{4} = \frac{81}{16} = 5 \frac{1}{16}$

7).

$( - \frac{3}{4})^{3} = - \frac{27}{64}$

8).

$(2 \frac{1}{2})^{3} = ( \frac{5}{2})^{3} = \frac{125}{8} = 15 \frac{5}{8}$

Answer 2

Ответ:

$\Big(\dfrac{1}{2}\Big)^2=\dfrac{1}{4}\ \ ,\ \ \ \ \ \ \Big(-\dfrac{1}{2}\Big)^5=-\dfrac{1}{32}\ \ ,\ \ \ \ \ \ \Big(-\dfrac{2}{3}\Big)^2=\dfrac{4}{9}\ \ ,\\\\\\0,3^3=0,3\cdot 0,3\cdot 0,3=0,027\\\\\\0,1^3=0,1\cdot 0,1\cdot 0,1=0,001\\\\\Big(1\dfrac{1}{2}\Big)^4=\Big(\dfrac{3}{2}\Big)^4=\dfrac{3^4}{2^4}=\dfrac{81}{16}\\\\\\\Big(-\dfrac{3}{4}\Big)^3=-\dfrac{3^3}{4^3}=-\dfrac{27}{64}\\\\\\\Big(2\dfrac{1}{2}\Big)^3=\Big(\dfrac{5}{2}\Big)^3=\dfrac{5^3}{2^3}=\dfrac{125}{8}$