1.x ²(x ²+1)+x ²(x+3)+3x+3=0 2.2x ³+3x ²+2x+3 =0 3. x(2x-7)-4x+14=0

Đáp án:

 $1, x^2(x^2+1)+x^2(x+3)+3x+3=0$

$⇔ x^4 +x^2 +x^3 +3x^2 +3x+3=0$

$⇔x^4+x^3+4x^2 +3x+3=0$

$⇔(x^4+x^3+x^2)+(3x^2+3x+3)+0$

$⇔x^2(x^2+x+1)+3(x^2+x+1)=0$

$⇔(x^2+x+1)(x^2+3)=0$

Th1 : $x^2+x+1 =0$

$⇔x^2+2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}  =0$

$⇔(x+\dfrac{1}{2})^2 + \dfrac{3}{4}$ =0$

Vì $(x+\dfrac{1}{2})^2 ≥ 0 ⇔(x+\dfrac{1}{2})^2 +\dfrac{3}{4} > 0$ (vô nghiệm)

Th2 : $x^2+3=0$

Vì $x^2≥ 0 ⇔x^2+3 > 0 $ (vô nghiệm)

$2 ) 2x^3 +3x^2+2x+3 =0$

$⇔ x^2(2x+3)+(2x+3)=0$

$⇔(2x+3)(x^2+1)=0$

Th1 : $x^2+1= 0$

Vì $x^2≥ 0 ⇔x^2+1 >0$ (vô nghiệm)

Th2 : $2x+3=0$

$⇔2x=-3$

$⇔x = -\dfrac{3}{2}$

Vậy $x=-\dfrac{3}{2}$

$3) x(2x-7)-4x+14=0$

$⇔x(2x-7)-2(2x-7)=0$

$⇔(2x-7)(x-2)=0$

⇔\(\left[ \begin{array}{l}2x-7=0\\x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=2\end{array} \right.\) 

Vậy $\text{x ∈ {$\dfrac{7}{2} ; 2$ }}$

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