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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 8: a, 3x^3 – 6x^2 -6x +12 =0 b, 8x^3 -8x^2 – 4x + 1=0

Toán Lớp 8: a, 3x^3 – 6x^2 -6x +12 =0
b, 8x^3 -8x^2 – 4x + 1=0

1. Giải đáp +Lời giải và giải thích chi tiết:
a) 3x^3 – 6x^2 – 6x + 12 = 0
⇔ 3x^2(x – 2) – 6(x – 2) = 0
⇔ (3x^2 – 6)(x – 2) = 0
⇔ 3(x^2 – 2)(x – 2) = 0
⇔ 3(x – sqrt{2})(x + sqrt{2})(x – 2) = 0
$⇔\left[\begin{matrix} x – \sqrt{2} = 0\\x + \sqrt{2} = 0 \\ x – 2 = 0\end{matrix}\right.$
$⇔ \left[\begin{matrix} x = \sqrt{2}\\ x = -\sqrt{2} \\ x = 2\end{matrix}\right.$
Vậy x ∈ { \sqrt{2}; -\sqrt{2}; 2 }
b) 8x^3 – 8x^2 – 4x + 1 = 0
⇔ 8x^3 – 12x^2 + 2x + 4x^2 – 6x + 1 = 0
⇔ 2x(4x^2 – 6x + 1) + (4x^2 – 6x + 1) = 0
⇔ (2x + 1)(4x^2 – 6x + 1) = 0
Trường hợp 1: 2x + 1 = 0
⇔ 2x = -1
⇔ x = -1/2
Trường hợp 2: 4x^2 – 6x + 1 = 0
⇔ (x^2 – 3/2x + 1/4). 4 = 0
⇔ x^2 – 2. x. 3/4 + (3/4)^2 – (3/4)^2 + 1/4 =0
⇔ (x – 3/4)^2 – 5/16 = 0
⇔ (x – 3/4 – sqrt{5}/4 )( x – 3/4 + sqrt{5}/4 ) = 0
⇔ (x – (3 + sqrt{5})/4 )( x – (3 – sqrt{5})/4 ) = 0
$⇔ \left[\begin{matrix} x = \dfrac{3 + \sqrt{5}}{4}\\ x = \dfrac{3 – \sqrt{5}}{4}\end{matrix}\right.$
Vậy x ∈ { -1/2; (3 + sqrt{5})/4; (3 – sqrt{5})/4 }