(5x+1)(3-4x)+20x^2 при x=-5 (5x+1)(3-4x)+20x^2=

Question

Алгебра: (5x+1)(3-4x)+20x^2 при x=-5 (5x+1)(3-4x)+20x^2=

vladislava240203444 · Accepted Answer

Х^6-12x^4+48x²-16-48=0 x^6-12x^4+48x²-64=0 (x^6-64)-12x²(x²-4)=0 (x²)³-(4)³-12x²(x²-4)=0 (x²-4)(x^4+4x²+16)-12x²(x²-4)=0 (x²-4)(x^4+4x²+16-12x²)=0 (x²-4)(x^4-8x²+16)=0 (x²-4)(x²-4)²=0 (x²-4)³=0 ⇒ x²-4=0 (x-2)(x+2)=0 x-2=0 x=2 x+2=0 x=-2 ответ: х1=2 х2=-2...

MOGZ ответил

(5x+1)(3-4x)+20x^2 при x=-5
(5x+1)(3-4x)+20x^2=