Question

[Trần Đức Minh]

Giải phương trình:

\(3x^2-4x-4=0\)

Answer 1

[Nguyễn Thị Thương Hoài]

3\(x^2\) - 4\(x\) - 4 = 0

3(\(x^2\) - 2. \(\dfrac{2}{3}\)\(x\) + \(\dfrac{4}{9}\)) - \(\dfrac{16}{3}\) = 0

3.(\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{3}\)

   (\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{9}\) 

   \(\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{4}{3}\\x-\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}x=\dfrac{4}{3}+\dfrac{2}{3}\\x=-\dfrac{4}{3}+\dfrac{2}{3}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

S = { -\(\dfrac{2}{3}\); 2}

Answer 2

[Kiều Vũ Linh]

3x² - 4x - 4 = 0

⇔ 3x² - 6x + 2x - 4 = 0

⇔ (3x² - 6x) + (2x - 4) = 0

⇔ 3x(x - 2) + 2(x - 2) = 0

⇔ (x - 2)(3x + 2) = 0

⇔ x - 2 = 0 hoặc 3x + 2 = 0

*) x - 2 = 0

⇔ x = 2

*) 3x + 2 = 0

⇔ 3x = -2

⇔ x = -2/3

Vậy S = {-2/3; 2}