16x^2 -8x+1=4(x+3)(4x-1) 9x^2 +6x+1 = (3x+1)(x-2)

2 bình luận về “16x^2 -8x+1=4(x+3)(4x-1) 9x^2 +6x+1 = (3x+1)(x-2)”

1. 16x^2 -8x+1=4(x+3)(4x-1)

   ⇔ 16x^2 -8x +1= 4(4x^2 -x+12x-3)

   ⇔ 16x^2 -8x +1= 4(4x^2 +11x-3)

   ⇔ 16x^2 -8x +1= 16x^2 + 44x – 12

   ⇔ -52x=-13

   ⇔ x= 1/4

   Vậy x= 1/4

   $\\$

   9x^2 +6x+1= (3x+1)(x-2)

   ⇔ 9x^2 + 6x+1= 3x^2 -6x+x-2

   ⇔ 9x^2 +6x +1= 3x^2 -5x -2

   ⇔ 6x^2 +11x +3=0

   ⇔ 6x^2 + 2x + 9x +3=0

   ⇔ 2x(3x+1) + 3(3x+1)=0

   ⇔ (3x+1)(2x+3)=0

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

   ⇔ $\left[\begin{matrix} x=\dfrac{-1}{3}\\ x=\dfrac{-3}{2}\end{matrix}\right.$

   Vậy S={-1/3 ; -3/2}

   Trả lời
2. *   16x^2-8x+1=4(x+3)(4x-1)

   <=>16x^2-8x+1=16x^2+44x-12

   <=>-8x-44x=-12-1

   <=>-52x=-13

   <=>x=1/4

          Vậy x=1/4

   *     9x^2+6x+1=(3x+1)(x-2)

   <=>9x^2+6x+1=3x^2-5x-2

   <=>9x^2-3x^2+6x+5x+1+2=0

   <=>6x^2+11x+3=0

   <=>(3+2x)(1+3x)=0

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

   ⇔ $\left[\begin{matrix} x=\dfrac{-1}{3}\\ x=\dfrac{-3}{2}\end{matrix}\right.$

   Vậy x∈{-3/2;-1/3}

   Trả lời