a) \(4x^2-4x=-1\)
\(\Leftrightarrow4x\left(x-1\right)=-1\)
\(\Leftrightarrow4x=-1\) hoặc \(x-1=-1\)
\(\Leftrightarrow x=\dfrac{-1}{4}\) hoặc \(x=0\)
Vậy S={\(\dfrac{-1}{4};0\)}
\(\text{a) }4x^2-4x=-1\\ \Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\\ \Leftrightarrow\left(2x-1\right)^2=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }x=\dfrac{1}{2}\\ \)
\(\text{ b) }8x^3+12x^2+6x+1=0\\ \Leftrightarrow\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3=0\\ \Leftrightarrow\left(2x+1\right)^3=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x-\dfrac{1}{2}\\ \text{Vậy }x=-\dfrac{1}{2}\)