giải phương trình d,x^3+3x^2+4x+2=0 e,x^4+x^2+6x-8=0 g,(x^2+1)^2=4(2x-1)

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1. $d.\ x^3+3x^2+4x+2=0$

   $⇔(x^3+1)+(3x^2+3x)+(x+1)=0$

   $⇔(x+1)(x^2+2x+2)=0$

   $⇔(x+1)[(x+1)^2+1]=0$

   Vì $(x+1)^2≥0∀x⇒(x+1)^2+1>0∀x$

   $⇒x+1=0$

   $⇔x=-1$

   Vậy $S=\{-1\}$

   $e.\ x^4+x^2+6x-8=0$

   $⇔(x^4-x^3)+(x^3-x^2)+(2x^2-2x)+(8x-8)=0$

   $⇔(x-1)(x^3+x^2+2x+8)=0$

   $⇔(x-1)[(x^3+2x^2)-(x^2+2x)+(4x+8)]=0$

   $⇔(x-1)(x+2)(x^2-x+4)=0$

   $⇔(x-1)(x+2)[(x-\frac{1}{2})^2+\frac{15}{4}]=0$

   Vì $(x-\frac{1}{2})^2≥0∀x⇒(x-\frac{1}{2})^2+\frac{15}{4}>0∀x$

   $⇒(x-1)(x+2)=0$

   $⇔\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.⇔\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.$

   Vậy $S=\{1;-2\}$

   $g.\ (x^2+1)^2=4(2x-1)$

   $⇔x^4+2x^2-8x+5=0$

   $⇔(x^4-x^3)+(x^3-x^2)+(3x^2-3x)-(5x-5)=0$

   $⇔(x-1)(x^3+x^2+3x-5)=0$

   $⇔(x-1)[(x^3-1)+(x^2-1)+(3x-3)]=0$

   $⇔(x-1)^2(x^2+x+1+x+1+3)=0$

   $⇔(x-1)^2(x^2+2x+5)=0$

   $⇔(x-1)^2[(x+1)^2+4]=0$

   Vì $(x+1)^2≥0∀x⇒(x+1)^2+4>0∀x$

   $⇒x-1=0$

   $⇔x=1$

   Vậy $S=\{1\}$.

    

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