Question

7.9 daje naj, klasa 1 technikum, matematyka zadania a-h​

Answer 1

Verified answer

[tex]a)\ \ (2^{-3})^{-4}:2^6=2^{12}:2^6=2^{12-6}=2^6=64\\\\\\b)\ \ (5^{-2})^{3}:(5^{-3})^2=5^{-6}:5^{-6}=1\\\\\\c)\ \ (3^{-5})^0:(3^2)^{-1}=3^0:3^{-2}=3^{0-(-2)}=3^{0+2}=3^2=9\\\\\\d)\ \ (1\frac{1}{5})^{-2}:0,3^{-2}=(1\frac{1}{5}:0,3)^{-2}=(\frac{6}{5}:\frac{3}{10})^{-2}=(\frac{\not6^2}{\not5_{1}}\cdot\frac{\not10^2}{\not3_{1}})^{-2}=4^{-2}=(\frac{1}{4})^2=\frac{1}{16}[/tex]

[tex]e)\ \ (-0,1)^{-5}:0,1^{-4}=-0,1^{-5}:0,1^{-4}=-0,1^{-5-(-4)}=-0,1^{-5+4}=-0,1^{-1}\\\\=-(\frac{1}{10})^{-1}=-10^1=-10\\\\\\f)\ \ (-\frac{1}{5})^6:(0,2)^{-4}=(\frac{1}{5})^6:(\frac{2}{10})^{-4}=(\frac{1}{5})^6:(\frac{1}{5})^{-4}=(\frac{1}{5})^{6-(-4)}=(\frac{1}{5})^{6+4}=\\\\=(\frac{1}{5})^{10}\\\\\\g)\ \ (-0,25)^5:(\frac{1}{4})^3=-0,25^5:(\frac{1}{4})^3=-(\frac{25}{100})^5:(\frac{1}{4})^3=-(\frac{1}{4})^5:(\frac{1}{4})^3=\\\\=-(\frac{1}{4})^{5-3}=-(\frac{1}{4})^2=-\frac{1}{16}[/tex]

[tex]h)\ \ (1\frac{2}{5})^{-2}:1,4^{-3}=(\frac{7}{5})^{-2}:(\frac{14}{10})^{-3}=(\frac{7}{5})^{-2}:(\frac{7}{5})^{-3}=(\frac{7}{5})^{-2-(-3)}=(\frac{7}{5})^{-2+3}=\\=(\frac{7}{5})^1=\frac{7}{5}=1\frac{2}{5}[/tex]

Zastosowane wzory

[tex](a^m)^n=a^{m\cdot n}\\\\a^m:a^n=a^{m-n}\\\\a^n:b^n=(a:b)^n\\\\(\frac{a}{b})^{-n}=(\frac{b}{a})^n[/tex]