Đáp án:
$ a) 2(x+2)^2 -x³ - 8=0$
$⇔2(x²+4x +4) - x³-8)=0$
$⇔ 2x² +8x +8 - x³-8=0$
$⇔-x³ +2x² +8x =0$
$⇔ -x³ +4x² -2x² +8x =0$
$⇔ -x²(x-4)-2x(x-4)=0$
$⇔(x-4)(-x²-2x)=0$
$⇔x-4 =0⇔x=4$
$⇔ -x²-2x =0 ⇔ -x(x+2)=0 ⇔x =0 hoặc x = -2
Vậy ......
$f)(x-1)(x² +5x-2) -x³ +1 =0$
$⇔ x³ +5x² -2x -x² -5x +2 -x³ +1 =0$
$⇔4x² -7x+3 =0$
$⇔ 4x² -4x -3x +3 =0$
$⇔ 4x(x-1)-3(x-1)=0$
$⇔(x-1)(4x-3)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\4x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=\dfrac{3}{4}\end{array} \right.\)
Vậy...
$g) x² -3x +2 =0$
$⇔ x² -x -2x +2 =0$
$⇔x(x-1)-2(x-1)=0$
$⇔(x-1)(x-2)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy ...
$h) x³-8x²+21x-18=0$
$⇔ x³ - 2x² -6x² +12x + 9x -18 =0$
$⇔x²(x-2)-6x(x-2)+9(x-2)=0$
$⇔(x-2)(x²-6x+9)=0$
$⇔(x-2)=0⇔x =2$
$⇔(x²-6x+9) =0⇔ (x-3)² = 0⇔x =3$
Vậy......
$i) x^4 +x² +6x -8 =0$
$⇔ x^4 +x² +6x - 8=0$
$⇔ x^4 +2x³ -2x³ -4x² +5x² +10x - 4x - 8 =0$
$⇔ x³(x+2) -2x²(x+2) +5x(x+2) -4(x+2) =0$
$⇔ (x+2)(x³ -2x²+5x-4)=0$
$⇔(x+2) = 0⇔x = -2$
$⇔ (x³ -2x² +5x -4 ) =0$
$⇔ x³ -x² -x²+x+4x -4 =0$
$⇔ x²(x-1)-x(x-1)+4(x-1)=0$
$⇔(x-1)(x²-x+4)=0$
$⇔x² - 2 . x . \dfrac{1}{2} + \dfrac{1}{4} +\dfrac{15}{4} = 0$
$⇔ (x-\dfrac{1}{2})² + \dfrac{15}{4} =0$
Vì $(x-\dfrac{1}{2})²≥ 0$ Nên$ (x-\dfrac{1}{2})² + \dfrac{15}{4} > 0 $ (loại)
⇔x-1 = 0⇔ x= 1
Vậy x ∈ {1 ; -2 }
