Question

Scrieti sub forma de putere utilizând regulile de calcul cu puteri (a, x, y ≠ 0):
${3}^{3} \times {3}^{ - 1} = \\ {5}^{8} \div {5}^{4} \times {5}^{ - 2} = \\ ( {4}^{3} )^{ - 3} \\ {x}^{11} \times {x}^{12} \times {x}^{13} \times {x}^{ - 30} = \\ {x}^{5} \times {x}^{ - 3} = \\ {x}^{4} \div {x}^{ - 4} = \\ {a}^{6} \times {a}^{ - 2} \div {a}^{3} \times {a}^{ - 1} =$
[(a²) ¯⁴]³ =
$(\frac{5}{9} )^{3} \times (\frac{5}{9})^{ - 4} \times ( \frac{5}{9} )^{4} =$
$[( \frac{x}{y} )^{ - 4} ]^{ - 2} = \\ ( \frac{3a}{xy} )^{ - 1}] ^{3} .$

Answer 1

[tex] 2^{3} [/tex]·$3^{-1}$=3²=9
$5^{8}$:$5^{4}$·$5^{-2}$=$5^{8-4-2}$=5²=25
$(4^{3} )^{-3}$=$4^{-9}$
$x^{11}$·$x^{12}$·$x^{13}$·$x^{-30}$=$x^{11+12+13-30}$=$x^{6}$
$x^{5}$·$x^{-3}$=$x^{5-3}$=$x^{2}$
$x^{4}$:$x^{-4}$=$x^{4+4}$=$x^{8}$
$a^{6}$·$a^{-2}$:a³·$a^{-1}$=$a^{6-2-3-1}$=$a^{0}$=1
$[(a^{2}) ^{-4}]^{3}$=$a^{-2*4*3}$=$a^{-24}$
$(\frac{5}{9})^{3-4+4}$=$(\frac{5}{9})^{3}$
[tex] (\frac{x}{y}) ^{8} [/tex]
$(\frac{3a}{xy})^{-3}$=$( \frac{xy}{3a})^{3}$=x³y³/27a³