Odpowiedź:
[tex]a)\ \ 2^6\cdot(-2)^3:(-2)^8=2^6\cdot(-2)^3:2^8=-2^{6+3-8}=-2^1=-2\\\\\\b)\ \ \dfrac{5^7\cdot(-5)^2^4}{5^1^3\cdot5^1^8}=\dfrac{5^7\cdot5^2^4}{5^{13}\cdot5^{18}}=\dfrac{5^{7+24}}{5^{13+18}}=\dfrac{5^3^1}{5^3^1}=1\\\\\\c)\ \ \dfrac{(-7)^1^0}{7^8\cdot(-7)}=\dfrac{7^1^0}{-7^9}=-\dfrac{7^1^0}{7^9}=-7^{10-9}=-7^1=-7[/tex]
[tex]d)\ \ (-\frac{1}{4})^5:0,25^2=(-\frac{1}{4})^5:(\frac{1}{4})^2=-(\frac{1}{4})^{5-2}=-(\frac{1}{4})^3=-\frac{1}{64}\\\\\\e)\ \ (-0,1)^3\cdot0,1^4:0,1^2=-0,1^{3+4-2}=-0,1^5=-0,00001\\\\\\f)\ \ \dfrac{(-0,2)^3\cdot0,2^4\cdot(-0,2)^4}{-(-0,2^4\cdot0,2^6)}=\dfrac{(-0,2)^3\cdot0,2^4\cdot0,2^4}{-(-0,2^1^0)}=\dfrac{-0,2^{3+4+4}}{0,2^1^0}=\dfrac{-0,2^1^1}{0,2^1^0}=\\\\\\=-\dfrac{0,2^1^1}{0,2^1^0}=-0,2^{11-10}=-0,2^1=(-\frac{1}{5})^1=-\frac{1}{5}[/tex]
