Question

Помогите пожалуйста(

Answer 1

$\displaystyle\bf\\1)\\\\\frac{33^{n} }{3^{n-4}\cdot 11^{n} } =\frac{3^{n} \cdot 11^{n} }{3^{n-4}\cdot11^{n} }=\frac{3^{n} }{3^{n-4} } =3^{n-n+4} =3^{4} =81\\\\2)\\\\\frac{49^{n-1} }{7^{2n-3} } =\frac{(7^{2} )^{n-1} }{7^{2n-3} } =\frac{7^{2n-2} }{7^{2n-3} } =7^{2n-2-2n+3} =7^{1} =7\\\\3)\\\\\frac{a^{8n} b^{n-1} }{a^{2n} b^{n-2} } =a^{8n-2n} b^{n-1-n+2}=a^{6n} b\\\\4)\\\\\frac{x^{2n} +x^{-3n} }{x^{-n} } =\frac{x^{-n} \cdot(x^{3n} +x^{-2n} )}{x^{-n} } =x^{3n} +x^{-2n}$

$\displaystyle\bf\\5)\\\\\frac{4^{n+2} -4^{n} }{15} =\frac{4^{n} (4^{2} -1)}{15} =\frac{4^{n} \cdot 15}{15} =4^{n} \\\\6)\\\\\frac{3^{-n} +1}{3^{n} +1}=\frac{\dfrac{1}{3^{n} } +1}{3^{n} +1} =\frac{\dfrac{1+3^{n} }{3^{n} } }{3^{n} +1} =\frac{1+3^{n} }{3^{n} \cdot(3^{n} +1)} =\frac{1}{3^{n} }=3^{-n}$