Question

giải phương trình:

a,4x²-1=0

b,4x²+1=0

c,2x²-12x+17=0

Answer 1

a,4x²-1=0

<=>( 2x +1 ) . ( 2x -1 ) = 0

<=> 2x +1 = 0 hay 2x -1 = 0

<=> 2x = -1 hay 2x =1

<=> x = \(-\dfrac{1}{2}\) hay x = \(\dfrac{1}{2}\)

Vậy S = \(\left\{\dfrac{-1}{2};\dfrac{1}{2}\right\}\)

Answer 2

\(a)4x^2-1=0\\ \Leftrightarrow4x^2=1\\ \Leftrightarrow x^2=\dfrac{1}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

\(b)4x^2+1=0\\ \Leftrightarrow4x^2=-1\\ \Leftrightarrow x^2=-\dfrac{1}{4}(vôlí)\)

Vậy phương trình vô nghiệm