Question

tìm x 16x^2 - 1 =0

Answer 1

\(16x^2-1=0\\ \Rightarrow16x^2=0+1=1\\ \Rightarrow x^2=1:16=\dfrac{1}{16}\\ \Rightarrow x^2=\left(\dfrac{1}{4}\right)^2\\ \Rightarrow x=\dfrac{1}{4}\)

Answer 2

\(16x^2-1=0\)

\(16x^2=1\)

\(x^2=\dfrac{1}{16}\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Answer 3

`16x^2 -1 =0`

`(4x)^2 -1 =0`

`(4x-1)(4x+1) =0`

`<=>[(4x=1),(4x=-1):}`

`<=> [(x=1/4),(x=-1/4):}`

Vậy `x=+-1/4`