Question

Решите пожалуйста
Номер 5 пжж

Answer 1

$boxed{a}\\ 81x^3+18x^2+x=0\x*(81^2+18x+1)=0\x*(9x+1)^2=0\x=0\(9x+1)^2=0;;;;;;x=-frac{1}{9} \\ boxed{x_{1} =-frac{1}{9};;;x_{2} =0 }$

$boxed{b}\\2x^3*(x^2-16)=0\x*(x^2-16)=0\x=0\x^2-16=0;;;;x=4;;;;x=-4\\ boxed{x_{1}=-4 ;;;x_{2}=0 ;;;x_{3}=4}$