Błagam o pomoc
Klasa 7 proszę też gdybyście mogli o wytłumaczenie jak to się robi :/

[tex]a)\\\\2^{9} \cdot (\dfrac{1}{4} )^{9} \div (\dfrac{1}{2} )^{5}=(2\cdot \dfrac{1}{4} )^{9} \div (\dfrac{1}{2} )^{5}= ( \dfrac{1}{2} )^{9} \div (\dfrac{1}{2} )^{5}=(\dfrac{1}{2} )^{9-5} =(\dfrac{1}{2} )^{4}=\dfrac{1}{16} \\\\b)\\\\\dfrac{(\dfrac{1}{5} )^{8}\div (\dfrac{2}{25} )^{8} }{2,5^{6} } =\dfrac{(\dfrac{1}{5} \div \dfrac{2}{25} )^{8}}{(2\dfrac{1}{2} )^{6}} =\dfrac{(\dfrac{1}{5} \cdot \dfrac{25}{2} )^{8}}{(\dfrac{5}{2} )^{6}} =[/tex]

[tex]=\dfrac{( \dfrac{5}{2} )^{8}}{(\dfrac{5}{2} )^{6}}=( \dfrac{5}{2} )^{8-6}=( \dfrac{5}{2} )^{2}=\dfrac{25}{4} =6\dfrac{1}{4}[/tex]

korzystam ze wzorów:

[tex]x^{n} \cdot y^{n} =(x\cdot y)^{n} \\\\x^{n} \cdot x^{m} =(x)^{n+m} \\\\x^{n} \div x^{m} =(x)^{n-m} \\\\x^{n} \div y^{n} =(x\div y)^{n}[/tex]

Basetlago Basetlago · Answer 2

Korzystamy z własności potęg:

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

[tex]a) \ 2^{9}\cdot(\frac{1}{4})^{9}:(\frac{1}{2})^{5} =(2\cdot\frac{1}{4})^{9}:(\frac{1}{2})^{5} = (\frac{1}{2})^{9}:(\frac{1}{2})^{5} = (\frac{1}{2})^{9-5} = (\frac{1}{2})^{4} = \frac{1}{16}\\\\\\b) \ \frac{(\frac{1}{5})^{8}:(\frac{2}{25})^{8}}{2,5^{6}} = \frac{(\frac{1}{5}:\frac{25}{2})^{8}}{(\frac{25}{10})^{6}} =\frac{(\frac{1}{5}\cdot\frac{25}{2})^{8}}{(\frac{5}{2})^{6}} = \frac{(\frac{5}{2})^{8}}{(\frac{5}{2})^{6}} = (\frac{5}{2})^{8-6} = (\frac{5}{2})^{2} = \frac{25}{4} = 6\frac{1}{4}[/tex]