Szczegółowe wyjaśnienie:
Twierdzenia dotyczące potęg:
[tex]a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\\left(a\cdot b\right)^n=a^n\cdot b^n[/tex]
[tex]a)\ \left(5^{\sqrt3}\right)^{\sqrt3}=5^{\sqrt3\cdot\sqrt3}=5^3=125\\\\b)\ \left(3^{\sqrt2}\right)^{2\sqrt2}=3^{\sqrt2\cdot2\sqrt2}=3^{2\cdot2}=3^4=81\\\\c)\ \left(5^{\sqrt3-1}\right)^{\sqrt3+1}=5^{(\sqrt3-1)(\sqrt3+1)}=5^{(\sqrt3)^2-1^2}=5^{3-1}=5^2=25\\\\d)\ \left(2^{\sqrt7-\sqrt5}\right)^{\sqrt7+\sqrt2}=2^{(\sqrt7-\sqrt2)(\sqrt7+\sqrt2)}=2^{(\sqrt7)^2-(\sqrt2)^2}=2^{7-2}=2^5=32[/tex]
[tex]e)\ 7^{\sqrt2}\cdot49^{-\frac{\sqrt2}{2}}=7^{\sqrt2}\cdot\left(7^2\right)^{-\frac{\sqrt2}{2}}=7^{\sqrt2}\cdot7^{2\cdot\left(-\frac{\sqrt2}{2}\right)}\\=7^{\sqrt2}\cdot7^{-\sqrt2}=7^{\sqrt2+(-\sqrt2)}=7^0=1\\\\f)\ 9^{\sqrt5}\cdot3^{1-2\sqrt5}=\left(3^2\right)^{\sqrt5}\cdot3^{1-2\sqrt5}=3^{2\cdot\sqrt5}\cdot3^{1-2\sqrt5}=3^{2\sqrt5+1-2\sqrt5}=3^1=3\\\\g)\ \dfrac{2^{\sqrt3+6}}{2^{\sqrt3+1}}=2^{\sqrt3+6-(\sqrt3+1)}=2^{\sqrt3+6-\sqrt3-1}=2^5=32[/tex]
[tex]h)\ \dfrac{6^{\sqrt3+1}\cdot2^{-\sqrt3}}{3^{\sqrt3}}=\dfrac{(3\cdot2)^{\sqrt3+1}\cdot2^{-\sqrt3}}{3^{\sqrt3}}=\dfrac{3^{\sqrt3+1}\cdot2^{\sqrt3+1}\cdot2^{-\sqrt3}}{3^{\sqrt3}}\\\\=3^{\sqrt3+1-\sqrt3}\cdot2^{\sqrt3+1+(-\sqrt3)}=3^1\cdot2^1=3\cdot2=6[/tex]
