Question

помогите решить №1,2,3

Answer 1

№1 г)$frac{(m^{4})^{7}}{(m^3)^9*m}=frac{m^2^8}{m^2^8}=m^0=1$

 

№2

$(2^5)^2*3^1^0/6^T \2^1^0*3^1^0/6^T \6^1^0/6^T=6^{10-T}$</var></p> <p> </p> <p> </p> <p> </p> <p>№3</p> <p><img src=[/tex](3/5)^3*(5/3)^2=3/5 \1,6^0=1 \3/5<1" title="(2^5)^2*3^1^0/6^T \2^1^0*3^1^0/6^T \6^1^0/6^T=6^{10-T}" title="(3/5)^3*(5/3)^2=3/5 \1,6^0=1 \3/5<1" title="(2^5)^2*3^1^0/6^T \2^1^0*3^1^0/6^T \6^1^0/6^T=6^{10-T}" alt="(3/5)^3*(5/3)^2=3/5 \1,6^0=1 \3/5<1" title="(2^5)^2*3^1^0/6^T \2^1^0*3^1^0/6^T \6^1^0/6^T=6^{10-T}" />

 

 

 

№3

$(2^5)^2*3^1^0/6^T \2^1^0*3^1^0/6^T \6^1^0/6^T=6^{10-T}$

 

 

 

№3

[tex](3/5)^3*(5/3)^2=3/5 \1,6^0=1 \3/5<1" />

 

Answer 2

2)2^10*3^10/6T=6^20/6T=6^20-T