*) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
*)\(8x^3+12x^2+6x+1=0\Leftrightarrow8x^3+8x^2+2x+4x^2+4x+1\)
\(\Leftrightarrow2x\left(4x^2+4x+1\right)+\left(x^2+4x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)^2=0\Leftrightarrow\left(2x+1\right)^3=0\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
a,42 - 4x = -1
=> 4x2 - 4x + 1 = 0
=> (2x - 1)2 = 0
=> 2x -1 = 0
=> x = \(\dfrac{1}{2}\)
b, 8x3 +12x2 + 6x +1=0
=> ( 2x +1 )3 = 0
=> 2x + 1 = 0
=> x = \(\dfrac{-1}{2}\)
