Question

\(c,\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{16}\)

Answer 1

`(x+1/3)^2 = 1/16`

`(x+1/3)^2`\(=\left(\pm\dfrac{1}{4}\right)^2\)

`@TH1:`

`x+1/3=1/4`

`x=1/4-1/3`

`x=3/12-4/12`

`x=-1/12`

`@TH2:`

`x+1/3=-1/4`

`x=-1/4-1/3`

`x=-3/13-4/12`

`x=-7/12`

Vậy `x = {-7/12; -1/12}`

Answer 2

`c,`

`(x+1/3)^2=1/16`

`=>(x+1/3)^2=(1/4)^2`

`Th1`

`x+1/3=1/4`

`=>x=1/4-1/3`

`=>x= -1/12`

`Th2`

`x+1/3=-1/4`

`=>x=-1/4-1/3`

`=>x=-7/12`

Answer 3

`(x + 1/3)^2 = 1/16`

`=>` \(\left[{}\begin{matrix}\left(x+\dfrac{1}{3}\right)^2=\left(\dfrac{1}{4}\right)^2\\\left(x+\dfrac{1}{3}\right)^2=\left(-\dfrac{1}{4}\right)^2\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{4}\\x+\dfrac{1}{3}=-\dfrac{1}{4}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=-\dfrac{7}{12}\end{matrix}\right.\)

Answer 4

\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{16}\)

\(\left(x+\dfrac{1}{3}\right)^2=\left(\dfrac{1}{4}\right)^2\)

`=>`\(x+\dfrac{1}{3}=\pm\dfrac{1}{4}\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{3}\\x=\left(-\dfrac{1}{4}\right)-\dfrac{1}{3}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=-\dfrac{7}{12}\end{matrix}\right.\)