a) Mạn phép sửa đề :
x4 - 3x3 + 4x2 - 3x + 1 = 0
⇔ x4 - x3 - 2x3 + 2x2 + 2x2 - 2x - x + 1 = 0
⇔ x3( x - 1) - 2x2( x - 1) + 2x( x - 1) - ( x - 1) = 0
⇔ ( x - 1)( x3 - 2x2 + 2x - 1) = 0
⇔ ( x - 1)[ ( x - 1)(x2 + x + 1) - 2x( x - 1)] = 0
⇔ ( x - 1)2( x2 - x + 1) = 0
Do : x2 - x + 1 \(=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\text{≥}\dfrac{3}{4}>0\text{∀}x\)
⇔ ( x - 1)2 = 0
⇔ x = 1
Vậy,....
b) 6x4 - x3 - 7x2 + x + 1 = 0
⇔ 6x4 + 6x3 - 7x3 - 7x2 + x + 1 = 0
⇔ 6x3( x + 1) - 7x2( x + 1) + x + 1 = 0
⇔ ( x + 1)( 6x3 - 7x2 + 1 ) = 0
⇔ ( x + 1)( 6x3 - 6x2 - x2 + 1 ) = 0
⇔ ( x + 1)[ 6x2( x - 1) -( x + 1)( x - 1)] = 0
⇔ ( x + 1)2( 6x2 - x - 1) = 0
⇔ ( x + 1)2( 6x2 - 3x + 2x - 1) = 0
⇔( x + 1)2[ 3x( 2x - 1) + 2x - 1] = 0
⇔( x + 1)2( 2x - 1)( 3x + 1) = 0
⇔ x = -1 ; x = \(\dfrac{1}{2}\) hoặc : x = \(\dfrac{-1}{3}\)
Vậy,....
