Question

помогите пожалуйста ​

Answer 1

Решение и ответ:

$\displaystyle 1)\;\;\frac{{{{\left({\frac{1}{{25}}}\right)}^{-n}}}}{{{5^{2n-1}}}}=\frac{{{{\left({\frac{1}{{5^{2} }}}\right)}^{-n}}}}{{{5^{2n-1}}}}= \frac{{{5^{2n}}}}{{{5^{2n-1}}}}={5^{2n-(2n-1)}}={5^{2n-2n+1}}={5^1}=5$

$\displaystyle 2)\;\;\frac{{{{12}^n}}}{{{2^{2n-1}}\cdot{3^{n+1}}}}=\frac{{{{\left({4\cdot3}\right)}^n}}}{{{2^{2n-1}}\cdot{3^{n+1}}}}=\frac{{{4^n}\cdot{3^n}}}{{{2^{2n-1}}\cdot{3^{n+1}}}}=\frac{{{2^{2n}}\cdot{3^n}}}{{{2^{2n-1}}\cdot{3^{n+1}}}}={2^{2n-(2n-1)}}\cdot{3^{n-(n+1)}}={2^{2n-2n+1}}\cdot{3^{n-n-1}}={2^1}\cdot{3^{-1}}=\frac{2}{3}$

$\displaystyle 3)\;\;\frac{{{{45}^{n+1}}}}{{{3^{2n+1}}\cdot{5^n}}}=\frac{{{{\left({9\cdot 5}\right)}^{n+1}}}}{{{3^{2n+1}}\cdot{5^n}}}=\frac{{{{\left({{3^2}\cdot 5}\right)}^{n+1}}}}{{{3^{2n+1}}\cdot{5^n}}}=\frac{{{{\left({{3^2}}\right)}^{n+1}}\cdot{5^{n+1}}}}{{{3^{2n+1}}\cdot{5^n}}}=\frac{{{3^{2n+2}}\cdot{5^{n+1}}}}{{{3^{2n+1}}\cdot{5^n}}}={3^{2n+2-(2n+1)}}\cdot{5^{n+1-n}}={3^{2n+2-2n-1}}\cdot{5^{n+1-n}}={3^1}\cdot{5^1}=15$

$\displaystyle 4)\;\;\frac{{{{60}^n}}}{{{2^{2n}}\cdot{3^{n-1}}\cdot{5^{n+1}}}}=\frac{{{{\left({4\cdot3\cdot5}\right)}^n}}}{{{2^{2n}}\cdot{3^{n-1}}\cdot{5^{n+1}}}}=\frac{{{{\left({{2^2}\cdot3\cdot5}\right)}^n}}}{{{2^{2n}}\cdot{3^{n-1}}\cdot{5^{n+1}}}}=\frac{{{2^{2n}}\cdot{3^n}\cdot{5^n}}}{{{2^{2n}}\cdot{3^{n-1}}\cdot{5^{n+1}}}}={2^{2n-2n}}\cdot{3^{n-(n-1)}}\cdot{5^{n-(n+1)}}={2^0}\cdot{3^1}\cdot{5^{-1}}=\frac{{1\cdot 3}}{5}=\frac{3}{5}$