Question

me ajuda a 53 e a 54 por favor

Answer 1

53)$x=(2- \frac{1}{2})^-^2 |y=(1+ \frac{1}{3})^-^1$
X.Y=?
$x=(2- \frac{1}{2})^-^2$
realizando Minimo Multiplo Comum:
$x=( \frac{4}{2} - \frac{1}{2})^-^2$
$x=( \frac{3}{2} )^-^2$
potência com expoente negativo inverte
$x=\frac{2^2}{3^2}$
$x=\frac{4}{9}$
DE Y:
$y=(1+ \frac{1}{3})^-^1$
MMC:
$y=( \frac{3}{3} + \frac{1}{3})^-^1$
$y=( \frac{4}{3} )^-^1$
$y=\frac{3}{4}$
logo x.y=
$y=\frac{3}{4}|x=\frac{4}{9} -entao: \frac{3}{4}. \frac{4}{9}= \frac{12}{36}$
simplificando:
 $\frac{12}{36}/6 = \frac{2}{6}/2 =\frac{1}{3}$
54)$m=( \frac{4}{5}- \frac{1}{2} )^-^3|n=(3+ \frac{1}{3})^2$
M:
$m=( \frac{4}{5}- \frac{1}{2} )^-^3$
realizando o MMC
$m=( \frac{8}{10}- \frac{5}{10} )^-^3$
$m=( \frac{3}{10} )^-^3$
$m=( \frac{10}{3} )^3$
$m= \frac{1000}{27}$
Agora o valor de N:
$n=(3+ \frac{1}{3})^2$
Realizando o MMC:
$n=( \frac{9}{3} + \frac{1}{3})^2$
$n=( \frac{10}{3})^2$
$n= \frac{100}{9}$
m:n é:

$n= \frac{100}{9}|m= \frac{1000}{27}-\ \textgreater \ entao: \frac{ \frac{1000}{27} }{ \frac{100}{9} }$
entao:
$\frac{ \frac{1000}{27} }{ \frac{100}{9} }= \frac{9000}{2700}$
simplificando:

$\frac{9000}{2700}/900=\frac{10}{3}$