C1

Ta có \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-6\right)\left(x-2\right)\)

C2 

ta có \(x^2-8x+12\)

\(=x^2-8x+16-4\)

\(=\left(x-4\right)^2-2^2\)

\(=\left(x-4-2\right)\left(x-4+2\right)\)

\(=\left(x-6\right)\left(x-2\right)\)

\(12x^2+7x-12=12x^2-5x+12x-12\)

\(=x\left(12x-5\right)+12\left(x-1\right)\)

Đề sai rồi bạn ời

a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)

b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)

d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)

e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)

\(=\left(x+4\right)^2\)

\(x^2+8x+16=\left(x+4\right)^2\)

3x2 + 8x + 2 = 0

Có a = 3; b' = 4; c = 2

⇒ Δ’ = 42 – 2.3 = 10 > 0

⇒ Phương trình có hai nghiệm phân biệt:

a) \(x^4+8x+63\)

\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)

\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)

\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)

c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)

Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)

\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)

\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)

\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)

\(2x^3-5x^2+8x-3\)

\(\Leftrightarrow2x^3-x^2-4x^2+2x+6x+3\)

\(\Leftrightarrow x^2\cdot\left(2x-1\right)-2x\cdot\left(2x-1\right)+3\cdot\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-1\right)\cdot\left(x^2-2x+3\right)\)

\(2x^3-5x^2+8x-3\)

\(=2x^3-x^2-4x^2+2x+6x-3\)

\(=x^2\left(2x-1\right)-2x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(2x-1\right)\left(x^2-2x+3\right)\)

 = 2x^3  - 4x^2 - x^2 + 2x + 6x - 3 

 = 2x^2 ( x - 1/2 ) - x ( x - 1/2 ) +3 ( x - 1/2 )

= ( x - 1/2 )( 2x^2 - x + 3 )