Responda: (todos os numeros depois dos parenteses significam cubo ou quadrado) (-1/2)4= (-2/7)0= (-1.... Pergunta de ideia deAndrei7Ano

robertocarlos5otivr9 $\left(-\dfrac{1}{2}\right)^4=\left(\dfrac{1}{2}\right)^4=\dfrac{1^4}{2^4}=\dfrac{1}{16}$

$\left(-\dfrac{2}{7}\right)^{0}=1$ (todo número elevado a zero é igual a 1)

$\left(-\dfrac{127}{538}\right)^{1}=-\dfrac{127}{538}$ (todo número elevado a 1 é igual a ele mesmo)

$\left(\dfrac{2}{3}\right)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}$

$\left(2\dfrac{1}{4}\right)^{2}=\left(\dfrac{2\cdot4+1}{4}\right)^{2}=\left(\dfrac{8+1}{4}\right)^2=\left(\dfrac{9}{4}\right)^{2}=\dfrac{9^2}{4^2}=\dfrac{81}{16}$

$(0,4)^{3}=\left(\dfrac{4}{10}\right)^{3}=\left(\dfrac{2}{5}\right)^{3}=\dfrac{2^3}{5^3}=\dfrac{8}{125}$

$\left(-3\dfrac{2}{5}\right)^{1}=\left(\dfrac{(-3)\cdot5+2}{5}\right)^{1}=\left(\dfrac{-15+2}{2}\right)^{1}=\left(-\dfrac{13}{2}\right)^{1}=-\dfrac{13}{2}$

$\sqrt{\dfrac{1}{16}}=\dfrac{\sqrt{1}}{\sqrt{16}}=\dfrac{1}{4}$

$-\sqrt{\dfrac{64}{81}}=-\dfrac{\sqrt{64}}{\sqrt{81}}=-\dfrac{8}{9}$

$\sqrt{2\dfrac{1}{4}}=\sqrt{\dfrac{2\cdot4+1}{4}}=\sqrt{\dfrac{8+1}{4}}=\sqrt{\dfrac{9}{4}}=\dfrac{\sqrt{9}}{\sqrt{4}}=\dfrac{3}{2}$

$-\sqrt{\dfrac{49}{100}}=-\dfrac{\sqrt{49}}{\sqrt{100}}=-\dfrac{7}{10}$

1 votes Thanks 2

ivanildoleiteba Olá, boa noite ☺

Resoluções:

a)

(-1/2)⁴ = (-1/2).(-1/2).(-1/2).(-1/2)=1/16.

b)

(-2/7)⁰=1.

c)

(-127/538)¹=-127/538

d)

(2/3)²=(2/3).(2/3)=4/9

e)

$\\ (2 \dfrac{1}{4} )= \dfrac{9}{4}
\\
\\ (\dfrac{9}{4})^2=(\dfrac{9}{4}) \cdot (\dfrac{9}{4})= \dfrac{81}{16}$

f)

(0,4)³=(0,4).(0,4).(0,4)=0,064.

g)

$(-3 \dfrac{2}{5} )^1=-3 \dfrac{2}{5}$

h)

√1/16=1/16.

i)

-√64/81=-(√(8²·8²))/81=-8/81

j)

$\sqrt{2 \dfrac{1}{4} } = \sqrt{ \dfrac{9}{4} } = \dfrac{3}{2}$

k)

-√49/100=-(√(7²))/100=-7/100.

Bons estudos :)

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