Toán Lớp 8: (x-2)^3-3(x+1)(x-1)+6x^2=5 Help mik nha :3

Question

Toán Lớp 8:  (x-2)^3-3(x+1)(x-1)+6x^2=5 Help mik nha :3

landinhngoc97 · Answer

Giải đáp+Lời giải và giải thích chi tiết: $ $ (x-2)^3-3(x+1)(x-1)+6x^2=5 $ $ <=>x^3-6x^2+12x-8-3(x^2-1)+6x^2=5 $ $ <=>x^3-6x^2+6x^2+12x-8-3x^2+3-5=0 $ $ <=>x^3-3x^2+12x-10=0 $ $ <=>x^3-x^2-2x^2+2x+10x-10=0 $ $ <=>x^2(x-1)-2x(x-1)+10(x-1)=0 $ $ <=>(x^2-2x+10)(x-1)=0 $ $ <=> ( left[ begin{array}{l}x-1=0 x^2-2x+10=0 end{array} right. ) $ $ <=> ( left[ begin{array}{l}x=1 x^2-2x+1+9=0 end{array} right. ) $ $ <=> ( left[ begin{array}{l}x=1(TM) (x-1)^2=-9(VôLý) end{array} right. ) $ $ text{Vậy}x=1

ankim · Answer

Answ  er    (x - 2)^3 - 3 . (x + 1) . (x - 1) + 6x^2 = 5   => x^3 - 6x^2 + 12x - 8 + 3 - 3x^2 + 6x^2 = 5   => x^3 + (6x^2 - 3x^2 - 6x^2) + 12x + (3 - 8) = 5   => x^3 + (3x^2 - 6x^2) + 12x - 5 = 5   => x^3 - 3x^2 + 12x - 5 = 5   => x^3 - 3x^2 + 12x - 5 - 5 = 0   => x^3 - 3x^2 + 12x - 10 = 0   => (x - 1) . (x^2 - 2x + 10) = 0   => $ left[ begin{matrix} x - 1 = 0   x^2 - 2x + 10 = 0 end{matrix} right.$   => $ left[ begin{matrix} x = 0 + 1   x . (x - 2) = 0 - 10 end{matrix} right.$   => $ left[ begin{matrix} x = 1    text{(TM)}   x . (x - 2) = - 10    text{(KTM)} end{matrix} right.$   $ text{Vậy}$ x = 1