Question

Answer 1

a)

\(4x^2-4x=-1\\ \Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x\right)^2-2.2x+1=0\\ \Leftrightarrow\left(2x-1\right)^2=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Vậy `x=1/2`

b)

\(8x^3+12x^2+6x+1=0\\ \Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1=0\\ \Leftrightarrow\left(2x+1\right)^3=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\)

Vậy `x=-1/2`

`HaNa♬D`

Answer 2

a: 4x^2-4x=-1

=>4x^2-4x+1=0

=>(2x-1)^2=0

=>2x-1=0

=>x=1/2

b: 8x^3+12x^2+6x+1=0

=>(2x+1)^3=0

=>2x+1=0

=>x=-1/2