a/( x+3) mũ3 – x ( 3x+1) mũ2 + ( 2x + 1) × ( 4x mũ2 – 2x + 1 ) = 28 b/ ( xã mũ2 – 1) mũ3 – ( xã mũ4 + xã mũ2 +1 ) × ( xã mũ2 – 1 ) = 0

0 bình luận về “a/( x+3) mũ3 – x ( 3x+1) mũ2 + ( 2x + 1) × ( 4x mũ2 – 2x + 1 ) = 28 b/ ( xã mũ2 – 1) mũ3 – ( xã mũ4 + xã mũ2 +1 ) × ( xã mũ2 – 1 ) = 0”

1. Đáp án:

    

   Giải thích các bước giải:

   $a,(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28$

   $⇔x^3+9x^2+27x+27-x(9x^2+6x+1)+(8x^3+1)=28$

   $⇔x^3+9x^2+27x-9x^3-6x^2-x+8x^3+1=28-27$

   $⇔(x^3-9x^3+8x^3)+(9x^2-6x^2)+(27x-x)=1-1$

   $⇔3x^2+26x=0$

   $⇔x(3x+26)=0$

   ⇔\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=0\\x=-\dfrac{26}{3}\end{array} \right.\) 

   $b,(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0$

   $⇔x^6-3x^4+3x^2-1-(x^6-1)=0$

   $⇔x^6-3x^4+3x^2-1-x^6+1=0$

   $⇔(x^6-x^6)-3x^4+3x^2=0$

   $⇔3x^2-3x^4=0$

   $⇔3x^2(1-x^2)=0$

   ⇔\(\left[ \begin{array}{l}3x^2=0\\1-x^2=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\) 

    

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