porownaj liczby

[wirus1910]

Porownaj liczby \(\displaystyle{ a ^{b}oraz b ^{a}}\)gdzie a=\(\displaystyle{ [ (2- \sqrt{3} )^{ \frac{1}{2} }+ (2+ \sqrt{3}) ^{ \frac{1}{2} } ]^{2}}\),
b=\(\displaystyle{ \frac{ 81^{-1}* \sqrt{3} }{ 27 ^{-2} * \sqrt[4]{9} }}\).

[sea_of_tears]

\(\displaystyle{ a : \newline
[(2-\sqrt3)^{\frac{1}{2}}+(2+\sqrt3)^{\frac{1}{2}}]^2=
[\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}]^2=
\sqrt{2-\sqrt3}^2+2\sqrt{2-\sqrt3}\sqrt{2+\sqrt3}+\sqrt{2+\sqrt3}^2=\newline
=2-\sqrt3+2\sqrt{(2-\sqrt3)(2+\sqrt3)}+2+\sqrt3=
4+2\sqrt{1}=4+2=6
\newline
\newline
b :\newline
\frac{81^{-1}\cdot \sqrt3}{27^{-2}\cdot \sqrt[4]{9}}=
\frac{(3^4)^{-1}\cdot 3^{\frac{1}{2}}}{(3^3)^{-2}\cdot (3^2)^{\frac{1}{4}}}=
\frac{3^{-4}\cdot 3^{\frac{1}{2}}}{3^{-6}\cdot 3^{\frac{1}{2}}}=
\newline
=\frac{3^{-3\frac{1}{2}}}{3^{-5\frac{1}{2}}}=
3^2=9\newline
a^b=6^9\newline
b^a=9^6\newline
6^9>9^6}\)
bo :
\(\displaystyle{ 6^9>9^6\newline
3^9 \cdot 2^9 > 3^6 \cdot 3^6\newline
2^9>3^3\newline
512>27}\)