Đáp án:
1)
$x^{4}$ + $x^{2}$ + 6x - 8 = 0
<=> $x^{4}$ - $x^{2}$ + 2.$x^{2}$ -2x + 8x - 8 = 0
<=> x²(x²-1) + 2x(x-1) + 8(x-1) =0
<=> x²(x-1)(x+1) + 2x(x-1) + 8(x-1) =0
<=> (x-1).[x².(x+1) + 2x+8] = 0
<=> (x-1).[x³+x²+2x+8] = 0
<=> \(\left[ \begin{array}{l}x-1=0=> x =1\\x³+x²+2x+8=> x=-2 \end{array} \right.\)
Vậy x = 1 hoặc x = -2
3)
x³ + 3x² + 4x + 2= 0
<=> x³ + x² + 2x² + 2x + 2x + 2 = 0
<=> x²(x+1) + 2x(x+1) +2(x+1) = 0
<=> (x+1)( x² + 2x + 2 ) = 0
=> x = -1 (vì x² + 2x + 2 > 0 ∀x )
Vậy x = -1
