a) x4 + 2x3 + 5x2 + 4x - 12 = 0
=> x4 - x3+ 3x3- 3x2 + 8x2 -8x + 12x - 12 = 0
=> x3( x - 1) + 3x2( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0
=> ( x - 1)( x3 + 3x2 + 8x + 12 ) = 0
=> ( x - 1)( x3 + 2x2 + x2 + 2x + 6x + 12 ) = 0
=> ( x - 1)[ x2( x + 2) + x( x + 2) + 6( x + 2) ] = 0
=> ( x - 1)( x + 2)( x2 + x + 6 ) = 0
Ta thấy : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)
=> ( x - 1)( x + 2 ) = 0
=> x = 1 hoặc x = -2
Vậy,-
b) ( x + 1)3 + ( x - 2)3 = ( 2x - 1)3
=>x3+ 3x2 + 3x + 1 + x3 - 6x2 + 12x - 8 - ( 8x3 - 12x2 + 6x - 1)=0
=> 2x3 - 3x2 + 15x - 7 - 8x3 + 12x2 - 6x + 1 = 0
=> 9x2 - 6x3 + 9x - 6 = 0
=> 9x( x + 1) -6( x3 + 1 ) = 0
=> 9x( x + 1) - 6( x + 1)( x2 - x + 1) = 0
=> 3( x + 1)( 3x - 2x2 + 2x - 2) = 0
=> 3( x + 1)( - 2x2 + 5x - 2) = 0
=> 3( x + 1)( - 2x2 + x + 4x - 2) = 0
=> 3( x + 1)[ x( 1 - 2x ) - 2( 1 - 2x ) ] = 0
=> 3( x + 1)( 1 - 2x )( x - 2) = 0
Suy ra :
* x + 1 = 0 => x = -1
* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)
* x - 2 = 0 => x = 2
Vậy,-.
