Question

Tìm x: x^3 - 8 - (x - 2)(x - 12) = 0

Answer 1

x³ - 8 - (x - 2).(x - 12) = 0

<=> x³ - 2³ - (x - 2).(x - 12) = 0

<=> (x - 2).(x² + 2x + 4) - (x - 2).(x - 12) = 0

<=> (x - 2).(x² + 2x + 4 - x + 12) = 0

<=> (x - 2).(x² + x + 16) = 0

<=> x - 2 = 0

<=> x = 2

Vậy: x = 2

Answer 2

x³ - 8 - ( x - 2 )( x - 12 ) = 0

⇔ ( x - 2 )( x² + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0

⇔ ( x - 2 )( x² + 2x + 4 - x + 12 ) = 0

⇔ ( x - 2 )( x² + x + 16 ) = 0

⇔ x - 2 = 0 hoặc x² + x + 16 = 0

⇔ x = 2 < do x² + x + 16 = ( x² + x + 1/4 ) + 63/4 = ( x + 1/2 )² + 63/4 ≥ 63/4 > 0 ∀ x >

Answer 3

\(x^3-8-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\)

\(\Leftrightarrow x-2=0\) ( vì \(x^2+x+16>0\forall x\) )

\(\Leftrightarrow x=2\)

vậy.......