Question

(9а²+3а+1)(3а-1) если а=1/3

Answer 1

Объяснение:

$(9a^2+3a+1)*(3a-1)=(3a-1)*((3a)^2+3a*1+1^2)=(3a)^3-1^3=27a^3-1$

$\displaystyle\\a=\frac{1}{3} \ \ \ \ \ \ \Rightarrow\\27*(\frac{1}{3} )^3-1=27*\frac{1}{27} -1=\frac{27}{27}-1=1-1=0.$

Answer 2

Ответ:

(9а²+3а+1)(3а-1) при а =1/3

(9×(1/3)2+3×1/3+1)(3×1/3-1)=(9×1/9+1+1)(1-1)=3×0=0