Question

Zapisz w najprostszej postaci i oblicz.

a) [tex](1,5)^{7}:(1,5)^{5}[/tex]

b) [tex](\frac{4}{3}) ^{4} x 3^{4}[/tex]

c) [tex](\frac{5}{7})^{5} x (1\frac{2}{5})^{6}[/tex]

d) [tex]\frac{49^{3} x 7^{3} }{7^{8} }[/tex]
x- razy
Z góry dziękuję <3

Answer 1

Verified answer

[tex]a)\ \ (1,5)^7:(1,5)^5=(1,5)^{7-5}=1,5^2=(\frac{15}{10})^2=(\frac{3}{2})^2=\frac{9}{4}=2 \frac{1}{4}\\\\b)\ \ (\frac{4}{3})^4\cdot3^4=(\frac{4}{\not3_{1}}\cdot\not3^1)^4=4^4=256\\\\c)\ \ (\frac{5}{7})^5\cdot(1\frac{2}{5})^6=(\frac{5}{7})^5\cdot(\frac{7}{5})^6=(\frac{5}{7})^5\cdot(\frac{5}{7})^{-6}=(\frac{5}{7})^{5-6}=(\frac{5}{7})^{-1}=(\frac{7}{5})^1=\frac{7}{5}=1\frac{2}{5}[/tex]

[tex]d)\ \ \dfrac{49^3\cdot7^3}{7^8}=\dfrac{(7^2)^3\cdot7^3}{7^8}=\dfrac{7^6\cdot7^3}{7^3}=\dfrac{7^{6+3}}{7^8}=\dfrac{7^9}{7^8}=7^{9-8}=7^1=7\\\\\\Wykorzystano\ \ wlasno\'sci\ \ poteg\\\\a^m\cdot a^n=a^{m+n}\\\\a^n\cdot b^n=(a\cdot b)^n\\\\(a^m)^n=a^{m\cdot n}\\\\\frac{a^m}{a^n}=a^{m-n}[/tex]