\(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
<=>\(\left(4x-1\right)^2=4\left(x+3\right)\left(4x-1\right)\)
<=>\(4x-1=4\left(x+3\right)\)
<=>\(4x-1=4x+12\)
<=> \(0x=13\) => Phương trình vô nghiệm
=> S = {∅}
16x2 - 8x + 1 = 4(x + 3)(4x -1)
<=> (4x)2 - 8x + 1 = (4x + 12)(4x -1)
<=> (4x - 1)2 = (4x + 12)(4x -1)
<=> (4x - 1)2 - (4x + 12)(4x -1) = 0
<=> (4x - 1)(4x - 1 - 4x - 12) = 0
<=> -13(4x - 1) = 0
<=> 4x - 1 = 0
<=> \(x=\frac{1}{4}\)
Mình không chắc, sai gì thì thông cảm hey~
16x2 - 8x + 1 = 4(x + 3)(4x - 1)
\(\Leftrightarrow\) 16x2 - 8x + 1 = 16x2 + 44x − 12
\(\Leftrightarrow\) (16x2 - 16x2) + (- 8x - 44x) = - 12 - 1
\(\Leftrightarrow\) - 52x = - 13
\(\Leftrightarrow\) x = \(\frac{-13}{-52}\)
\(\Leftrightarrow\) x = \(\frac{1}{4}\)
Vậy S = {\(\frac{1}{4}\)}
Chúc bạn học tốt!