(х+1)²(х²+2х)=12 (х²+3х+1)(х²+3х+3)=-1 (х²-4х+1)(х²-4х+2)=12 - вопрос №1212313314142122238029 от hac20 29.06.2021 05:40

Question

Алгебра: (х+1)²(х²+2х)=12 (х²+3х+1)(х²+3х+3)=-1 (х²-4х+1)(х²-4х+2)=12

hac20 · Answer

1) (x+1)² (x² + 2x) = 12(x² + 2x + 1) (x² + 2x) = 12x² + 2x = t(t+1) t = 12t² + t - 12 = 0D = 1+48 = 49t₁ = 3t₂ = -4x² + 2x = 3x² + 2x - 3 = 0x₁ = 1x₂ = -3x² + 2x = -4x² + 2x + 4 =0D = 4-16 0нет корнейответ: 1; -32) x²  +3x = t(t+1) (t+3) = -1t² + 3t + t + 3 = -1t² + 4t + 4 = 0D₁ = 4-4 = 0t = -2x²  +3x = -2x² + 3x + 2=0x₁ = -1x₂ = -2ответ: -1; -23) x² - 4x = t(t+1) (t+2) = 12t² + 2t + t + 2 = 12t² + 3t - 10 = 0D = 9 + 40 = 49t₁ = (-3+7) / 2 = 2t₂ = (-3-7) / 2 = -5x² - 4x = 2x² - 4x - 2 = 0D₁ = 4+2...