Question

Решите уравнение:

б) (2x+1)(3x^2+1)(4x+1)=0

в)(x-8)(x^2-7x-8)=x^3-8x^2

 

 

 

 

Answer 1

a)$(2x+1)(3x^2+1)(4x+1)=0\ 2x+1=0 \ 2x=-1\ x=-frac{1}{2}\ 3x^2+1=0\ 3x^2=-1\ x^2=-frac{1}{3}\ x=+-sqrt{frac{1}{3}}\ 4x+1=0\ 4x=-1\ x=-frac{1}{4}$

б)$(x-8)(x^2-7x-8)=x^3-8x^2\ (x-8)(x+1)(x-8)=x^2(x-8)\ (x-8)^2(x+1)-x^2(x-8)=0\ (x-8)((x-8)(x+1)-x^2)=0\ (x-8)(x^2-7x-8-x^2)=0\ (x-8)(-7x-8)=0\ x-8=0\ x=8\ -7x-8=0\ -7x=8\ x=-1frac{1}{8}$