Odpowiedź:
[tex]Zad\ \ 5\\\\a)\ \ 4^2\cdot12^2\cdot(\frac{1}{4})^2=(\not4^1\cdot12\cdot\frac{1}{\not4_{1} })^2=12^2=144\\\\\\b)\ \ (-1,5)^3\cdot(-1\frac{1}{3})^3\cdot(-2)^3=-1,5 ^3\cdot(-\frac{4}{3})^3\cdot(-2)^3=-(1,5\cdot\frac{4}{3}\cdot2)^3=\\\\=-(\frac{15}{10}\cdot\frac{4}{3}\cdot2)^3=-(\frac{\not3^1}{\not2_{1}}\cdot\frac{\not4^2}{\not3_{1}}\cdot2)^3=-(2\cdot2)^3=-4^3=-64[/tex]
[tex]Zad.6\\\\a)\ \ (\frac{1}{7})^2\cdot14^2-10^5\cdot(\frac{1}{5})^5=(\frac{1}{\not7_{1} }\cdot\not14^2)^2-(\not10^2\cdot\frac{1}{\not5_{1}})^5=2^2-2^5=4-32=-28\\\\\\b)\ \ 35^2:(-7)^2+(-6)^3:(-2)^3=35^2:7^2+6^3:2^3=(35:7)^2+(6:2)^3=\\\\=5^2+3^3=25+27=52[/tex]
[tex]c)\ \ 4^2\cdot(\frac{1}{4})^2-(\frac{1}{5})^3\cdot5^3=(\not4\cdot\frac{1}{\not4})^2-(\frac{1}{\not5}\cdot\not5)^3=1^2-1^3=1-1=0\\\\\\d)\ \ (\frac{3}{7})^3:(-\frac{6}{7})^3-15^4:(-5)^4=-(\frac{3}{7}:\frac{6}{7})^3-15^4:5^4=-(\frac{\not3^1}{\not7_{1} }\cdot\frac{\not7^1}{\not6_{2}})^3-(15:5)^4=\\\\=-(\frac{1}{2})^3-3^4=-\frac{1}{8}-81=-81\frac{1}{8}[/tex]
