Question

8.Calculati:
e) $(1,2^{2} ) ^{-1}$
f) $[(-0,3)^{3} ] ^{-1}$
g) $[(0,1)^{-1} ] ^{-4}$
h) $[(1 \frac{1}{3} ) ^{2} ] ^{-1}$

Ma poate ajuta cineva?

Answer 1

$\displaystyle e).(1,2^2)^{-1}=(1,44)^{-1}=\left( \frac{144}{10} \right)^{-1}= \frac{10}{144}= \frac{5}{72}$

$\displaystyle f).[(-0,3)^3]^{-1}=(-0,027)^{-1}=\left(- \frac{27}{1000} \right)^{-1}=- \frac{1000}{27}$

$\displaystyle g).[(0,1)^{-1}]^{-4}=\left[\left( \frac{1}{10} \right)^{-1}\right]^{-4}=10^{-4}= \frac{1}{10^4} = \frac{1}{10000}$

$\displaystyle h). \left[\left(1 \frac{1}{3} \right)^2 \right]^{-1}=\left[\left( \frac{4}{3} \right)^2\right]^{-1}=\left( \frac{4^2}{3^2} \right)^{-1}=\left( \frac{16}{9} \right)^{-1}= \frac{9}{16}$

Answer 2

$( a^{ m})^{n} = a^{m*n}$

e).  $[(1.2)^{2}] ^{-1}$=$(1,2)^{-2}$=$\frac{1}{ (1,2)^{2} } = \frac{1}{1.44} = 0,69(4)$


f).$[(0.3)^{3}] ^{-1}$=$(-0,3)^{-3}=\frac{1}{ (-0.3)^{3} } = \frac{1}{-0.027} = -37.(037)$


g).$[ (0,1)^{-1}]^{-4}$=$(0,1)^{4}=0,0001$


h).$[ (1 \frac{1}{3})^{2} ]^{-1}$=$( \frac{4}{3} )^{-2}=\frac{1}{ ( \frac{4}{3} )^{2} } = \frac{1}{ \frac{16}{9} } = \frac{1}{1}* \frac{9}{16}= \frac{9}{16} =0.5625$


( $1 \frac{1}{3} = \frac{1*3 + 1}{3} = \frac{4}{3}$  am introdus intregii in fractie )