Срочно!!! 1) 3a^2+12/10a÷a^4-16/3a-6 при а=-9 2)c^9-1/5c^2+5c+5÷4c^6+4c^3+4/10c+10

Question

Срочно!!! 1) 3a^2+12/10a÷a^4-16/3a-6 при а=-9
2)c^9-1/5c^2+5c+5÷4c^6+4c^3+4/10c+10

Answer 1

1
(3a^2+12)/10a÷(a^4-16)/(3a-6)=3(a²+4)/(10a)*3(a-2)/[(a-2)(a+2)(a²+4)]=
=9/[10a(a+2)] 
при а=-9
9/[-9*10*(-9+2)]=9/(90*7)=1/70

2
(c^9-1)/(5c^2+5c+5)÷(4c^6+4c^3+4)/(10c+10 )=
=(c³-1)(c^6+c³+1)/[5(c²+c+1)]*10(c+1)/[4(c^6+c²+1)]=2(c³-1)(c+1)/(c²+c+1)=
=2(c-1)(c²+c+1)(c+1)/(c²+c+1)=2(c²-1)

Comments

Второе не правильно!