Question

Помогите с номерами срочно !!
Заранее спасибо!!

Answer 1

Объяснение:

10.

$\frac{(b^2)^3*b^{3n-2}*b^{3n}}{b^{6n}*(b^2)^2}=\frac{b^6*b^{3n-2}*b^{3n}}{b^{6n}*b^4}=\frac{b^{6+3n-2+3n}}{b^{6n+4}}=\frac{b^{6n+4}}{b^{6n+4}}=1$

11.

$\frac{(c^2)^4*c^{5n-1}*c^3}{c^{5n+7}*c}=\frac{c^8*c^{5n-1}*c^3}{c^{5n+7}*c}=\frac{c^{8+5n-1+3}}{c^{5n+7+1}}=\frac{c^{5n+10}}{c^{5n+8}}=c^{5n+10-5n-8}=c^2$

12.

$\frac{(z^3)^2*z^{2n+5}*(z^0)^{25}}{z^{2n}*z^7}=\frac{z^6*z^{2n+5}*1^{25}}{z^{2n}*z^7}=\frac{z^{6+2n+5}}{z^{2n+7}}=\frac{z^{2n+11}}{z^{2n+7}}=z^{2n+11-2n-7}=z^4$

13.

$\frac{(x^2)^n*(x^3)^2*x^5}{x^{2n+3}*x^4}=\frac{x^{2n}*x^6*x^5}{x^{2n+3}*x^4}=\frac{x^{2n+6+5}}{x^{2n+3+4}}=\frac{x^{2n+11}}{x^{2n+7}}=x^{2n+11-2n-7}=x^4$

14.

$\frac{3^n*2^{2n}*(3^2)^6*2^5}{(2^n)^2*2^3*3^{n+11}}=\frac{3^n*2^{2n}*3^{12}*2^5}{2^{2n}*2^3*3^{n+11}}=\frac{3^{n+12}*2^{2n+5}}{2^{2n+3}*3^{n+11}}=2^{2n+5-2n-3}*3^{n+12-n-11}=2^2*3^1=4*3=12$

15.

$\frac{36^2*2^{n+1}*3^{2n}}{2^n*9^n}=\frac{(3*2)^2*2^{n+1}*3^{2n}}{2^n*(3^2)^n}=\frac{3^2*2^2*2^{n+1}*3^{2n}}{2^n*3^{2n}}=\frac{3^{2n+2}*2^{n+3}}{2^n*3^{2n}}=2^{n+3-n}*3^{2n+2-2n}=2^3*3^2=8*9=72$