Bài 1. Giải phương trình (4x-1)(3x+2)=16x^2-1

2 bình luận về “Bài 1. Giải phương trình (4x-1)(3x+2)=16x^2-1”

1. (4x – 1)(3x + 2) = 16x^2 – 1

   ⇔ (4x – 1)(3x + 2) = (4x – 1)(4x + 1)

   ⇔ (4x – 1)(3x + 2) – (4x – 1)(4x + 1) = 0

   ⇔ (4x – 1)[(3x + 2) – (4x + 1)] = 0

   ⇔ (4x – 1)(3x + 2 – 4x – 1) = 0

   ⇔ (4x – 1)(-x + 1) = 0

   ⇔ \(\left[ \begin{array}{l}4x – 1=0\\-x + 1=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x =\dfrac{1}{4}\\-x = -1\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x =\dfrac{1}{4}\\x = 1\end{array} \right.\) 

   Vậy S = {1/4 ; 1}

   Trả lời
2. (4x-1)(3x+2)=16x^2-1

   <=>(4x-1)(3x+2)=(4x-1)(4x+1)

   <=>(4x-1)(3x+2)-(4x-1)(4x+1)=0

   <=>(4x-1)(3x+2-4x-1)=0

   <=>(4x-1)(-x+1)=0

   TH1 : 4x-1=0

   <=>4x=1

   <=>x=1/4

   TH2 : -x+1=0

   <=>x=1

   vậy S={1/4 ; 1}

    

   Trả lời