1, x3 + 1 = ( x + 1 )( x - 5 )
<=> ( x + 1 )( x2 - x + 1 ) = ( x + 1 )( x - 5 )
<=> ( x + 1 )( x2 - 2x + 6 ) = 0
Mà x2 - 2x + 6 = ( x - 1 )2 + 5 > 0
=> x + 1 = 0 <=> x = - 1
=> S = { - 1 }
2, x3 + 3x2 + 6x + 4 = 0
<=> x3 + x2 + 2x2 + 2x + 4x + 4 = 0
<=> x2( x + 1 ) + 2x( x + 1 ) + 4( x + 1 ) = 0
<=> ( x + 1 )( x2 + 2x + 4 ) = 0
Mà x2 + 2x + 4 = ( x + 1 )2 + 3 > 0
=> x + 1 = 0 <=> x = - 1
=> S = { - 1 }
1, \(x^3+1=\left(x+1\right)\left(x-5\right)\)
\(\Leftrightarrow x^3+1=x^2+x-5x-5\)
\(\Leftrightarrow x^3+1=x^2-4x-5\)
\(\Leftrightarrow x^3+1-x^2+4x+5=0\)
\(\Leftrightarrow x^3-x^2+4x+6=0\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(2x^2+2x\right)+\left(6x+6\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)-2x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+6\right)=0\) (1)
Ta có:
\(x^2-2x+6=x^2+2.x.1+1+5\\ =\left(x+1\right)^2+5\ge5>0\forall x\)
PT(1) xảy ra khi: \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
2, \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\) (2)
Ta có:
\(x^2+2x+4=x^2+2.x.1+1+3\\ =\left(x+1\right)^2+3\ge3>0\forall x\)
PT(2) xảy ra khi: \(x+1=0\Leftrightarrow x=-1\)
Vậy pt có 1 nghiệm duy nhất là x = -1
