Tìm `x:` `(3x + 3)^2 + (4x^2 -4)^4 = 0`

1 bình luận về “Tìm `x:` `(3x + 3)^2 + (4x^2 -4)^4 = 0`”

1. Vì {((3x + 3)^2 \ge 0),((4x^2 – 4)^4 \ge 0):} => (3x + 3)^2 + (4x^2 – 4)^2 \ge 0

   Mà (3x +3 )^2 + (4x^2 – 4)^4 = 0

   => {((3x + 3)^2 = 0),((4x^2 – 4)^4 = 0):}

   => {(3x + 3 = 0),(4x^2 – 4 = 0):}

   => {(3x = -3),(4x^2 = 4):}

   => {(x = -1),(x^2 = 1):}

   => {(x = -1),(x = +-1):}

   => x = +-1

   Vậy x = +-1

   Trả lời