Question

помогите пожалуйста!!! даю 20 баллов срочно 2.19 и 2.21

Answer 1

I hope this helps you

Answer 2

$1)\; \; \frac{3^5+3^9}{3^{-5}+3^{-9}}=\frac{3^5\, (1+3^4)}{3^{-9}\, (3^4+1)}=3^{5+9}=3^{14}\\\\2)\; \; \frac{2^5+2^6+2^7}{2^{-5}+2^{-6}+2^{-7}}=\frac{2^5\, (1+2+2^2)}{2^{-7}\, (2^2+2+1)}=2^{5+7}=2^{12}$

$3)\; \; \frac{(x^2y^3z^2)^4\cdot (x^3y)^3}{(xy^2z^4)^2}=\frac{x^8y^{12}z^8\cdot x^9y^3}{x^2y^4z^8}=\frac{x^{17}y^{15}}{x^2y^4}=x^{15}y^{11}\\\\4)\; \; \frac{(3^{n}a^2b^{n+1})^2\cdot (ab)^{n}}{(3a^2b^{n})^2}=\frac{3^{2n}a^4\, b^{2n+2}\cdot a^{n}b^{n}}{3^2a^4\, b^{2n}}=3^{2n-2}\cdot a^{n}\cdot b^{n+2}$