Trang chủ » Hỏi đáp » Môn Toán Bài 1. Giải phương trình (4x-1)(3x+2)=16x^2-1 26/05/2023 Bài 1. Giải phương trình (4x-1)(3x+2)=16x^2-1
(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
(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
