\(\left(x-1\right)\left(x+2\right)\left(x^2+x-4\right)\)
\(\left[{}\begin{matrix}x=1\\x=-2\\vn\end{matrix}\right.\)
x4 + x2 +6x - 8 = 0
<=> x4 - x3 +x3 - x2 +2x2 - 2x +8x - 8 = 0
<=>(x4- x3) + (x3 - x2) + (2x2 - 2x) +(8x - 8) = 0
<=> x3(x - 1) + x2(x - 1) + 2x(x - 1) +8(x - 1) = 0
<=> (x - 1)(x3 + x2 +2x +8 ) = 0
=> x - 1 = 0 <=> x = 1
hoặc x3 +x2 +2x +8 = 0
<=> x3 + 2x2 - x2 - 2x + 4x +8 = 0
<=> (x3 + 2x2) - (x2 + 2x)+ (4x +8) = 0
<=> x2(x + 2) - x(x + 2) + 4(x + 2) = 0
<=> (x + 2)(x2 - x +4) = 0
=> x + 2 = 0 <=> x = -2
hoặc x2 - x + 4 = 0
<=> [x2 - 2.\(\dfrac{1}{2}\).x + (\(\dfrac{1}{2}\))2] - (\(\dfrac{1}{2}\))2 + 4 = 0
<=> (x - \(\dfrac{1}{2}\))2 +\(\dfrac{15}{4}\) = 0
ta có : (x - \(\dfrac{1}{2}\))2 \(\ge\) 0 (\(\forall\) x\(\in\) R)
=> (x +\(\dfrac{1}{2}\))2 + \(\dfrac{15}{4}\) \(\ge\) \(\dfrac{15}{4}\) (\(\forall\) x \(\in\) R )
Vậy phương trình có tập nghiệm S = {1 ; -2}