9x^2 -1=(3x+1)(2x-3) 27x^2(x+3)-12(x^2 +3x)=0

0 bình luận về “9x^2 -1=(3x+1)(2x-3) 27x^2(x+3)-12(x^2 +3x)=0”

1. 9x²-1=(3x+1)(2x-3)

   ⇔(3x-1)(3x+1)-(3x+1)(2x-3)=0

   ⇔(3x+1)(3x-1-2x+3)=0

   ⇔(3x+1)(x+2)=0

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

   Vậy S={-1/3;-2}

   27x²(x+3)-12(x² +3x)=0

   ⇔27x²(x+3)-12x(x+3)=0

   ⇔(x+3)(27x²-12x)=0

   ⇔(x+3)3x(9x-4)=0

   ⇔x+3=0; 3x=0 hoặc 9x-4=0

   ⇔x=-3; x=0; x=4/9

   Vậy S=Ơ-3;0;4/9}

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2. Đáp án:

    

   Giải thích các bước giải:

    a. $9x^{2}$ – 1 = ( 3x + 1 )( 2x – 3 )
   <=> ( 3x +1 )(3x + 1 ) – ( 3x + 1 )( 2x – 3 ) = 0
   <=> ( 3x + 1 )( 3x – 1 – 2x + 3 ) = 0
   <=> ( 3x + 1 ) ( x + 2 ) = 0
   <=> \(\left[ \begin{array}{l}3x + 1 = 0 \\x + 2 = 0\end{array} \right.\)
   <=> \(\left[ \begin{array}{l}x=\frac{-1}{3}\\x=-2\end{array} \right.\) 

   S = { $\frac{-1}{3}$ ; -2 }

   b. $27x^{2}$ ( x + 3 ) – 12 ( $x^{2}$ + 3x ) = 0
   <=> $27x^{2}$ ( x + 3 ) – 12x ( x + 3 ) = 0
   <=> ( x + 3 )( $27x^{2}$ – 12x ) = 0
   <=> 3x( x + 3 )( 9x + 4 ) = 0
   <=> 3x = 0 ; x + 3 = 0 và 9x + 4 = 0
   <=> x = 0 ; x = -3 và x = $\frac{-4}{9}$

   S = { 0; -3; $\frac{-4}{9}$ }

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