tìm x 5x^10-40x^2=0 (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0 giúp em vs ạ!

1 bình luận về “tìm x 5x^10-40x^2=0 (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0 giúp em vs ạ!”

1. 5x^10 – 40x^2 = 0

   <=> 5x^2 (x^8 – 8) = 0

   <=> [(x^2 = 0),(x^8 = 8):} <=> [(x = 0),(x = root(8)(8)):}

   $$

   $$

   $$

   (x – 2)(x^2 + 2x + 7)+ 2(x^2 – 4) – 5(x – 2) = 0

   <=> x^3 + 2x^2 + 7x – 2x^2 – 4x – 14 + 2x^2 – 8 – 5x + 10 = 0

   <=> x^3 + 2x^2 – 2x – 12 = 0

   <=> x^3 + 4x^2 + 6x – 2x^2 – 8x – 12 = 0

   <=> x(x^2 + 4x + 6) – 2(x^2 + 4x + 6) = 0

   <=> (x – 2)(x^2 + 4x + 6) = 0

   Do x^2 + 4x + 6 > x^2 + 4x + 4 >= 0 AA x

   <=> x – 2 = 0 <=> x = 2

    

   Trả lời