MegaMatma: Klasówka Porównywanie potęg.

$\{[(-\frac{1}{2})^2]^3\}^4=[(\frac{1}{4})^3]^4=(\frac{1}{64})^4=[(\frac{1}{8})^2]^4=(\frac{1}{8})^8$	$T$	$N$
$\{[(-\frac{1}{2})^2]^3\}^4=[(\frac{1}{4})^3]^4=(\frac{1}{16})^4=[(\frac{1}{8})^2]^4=(\frac{1}{8})^8$	$T$	$N$

$[(\frac{3}{4})^{-2}]^3 < (1 \frac{1}{5})^6,$ ponieważ $[(\frac{3}{4})^{-2}]^3=(\frac{3}{4})^6$ a $\frac{3}{4}<1 \frac{1}{5}$	$P$	$F$
$[(\frac{3}{4})^{-2}]^3 < (1 \frac{1}{5})^6,$ ponieważ $[(\frac{3}{4})^{-2}]^3=(\frac{3}{4})^{-6}=(\frac{4}{3})^6=(\frac{16}{15})^6$ oraz $(1 \frac{1}{5})^6=(\frac{6}{5})^6=(\frac{18}{15})^6$ a $\frac{16}{15}< \frac{18}{15}$	$P$	$F$
$[(\frac{3}{4})^{-2}]^3 > (1 \frac{1}{5})^6,$ ponieważ $[(\frac{3}{4})^{-2}]^3=(\frac{3}{4})^{-6}=(\frac{4}{3})^6=(1 \frac{1}{3})^6$ a $1 \frac{1}{3}>1 \frac{1}{5}$	$P$	$F$

3.7 Klasówka Porównywanie potęg.