tìm x (3x-5)^2-(1-2x)^2=0
Giải thích
(3x – 5)2 – (1 – 2x)2 = 0
⇔ (3x – 5 + 1 – 2x)(3x – 5 – 1 + 2x) = 0
⇔ (x – 4)(5x – 6) = 0
⇔ \(\left[ \begin{array}{l}x = 4\\x = \frac{6}{5}\end{array} \right.\)
(3x – 5)2 – (1 – 2x)2 = 0
⇔ (3x – 5 + 1 – 2x)(3x – 5 – 1 + 2x) = 0
⇔ (x – 4)(5x – 6) = 0
⇔ \(\left[ \begin{array}{l}x = 4\\x = \frac{6}{5}\end{array} \right.\)