Question

Tìm x biết: 

(4x - 1)² - 4(2x - 3)² - x - 4 = 0

Answer 1

\((4x-1)^2-4(2x-3)^2-x-4=0\\\Leftrightarrow (4x)^2-2\cdot4x\cdot1+1^2-4[(2x)^2-2\cdot2x\cdot3+3^2]-x-4=0\\\Leftrightarrow 16x^2-8x+1-4(4x^2-12x+9)-x-4=0\\\Leftrightarrow 16x^2-8x+1-16x^2+48x-36-x-4=0\\\Leftrightarrow (16x^2-16x^2)+(-8x+48x-x)+(1-36-4)=0\\\Leftrightarrow 39x-39=0\\\Leftrightarrow 39x=39\\\Leftrightarrow x=1\\Vậy:x=1\)

Answer 2

(4x - 1)² -4(2x - 3)² - x - 4 = 0

16x² - 8x + 1 - 4(4x² - 12x + 9) - x - 4 = 0

16x² - 8x + 1 - 16x² + 48x - 36 - x - 4 = 0

39x - 39 = 0

39x = 39

x = 39 : 39

x = 1