(x+1/4)^2=(1/8)^2

=>x+1/4=1/8

=> x=1/8-1/4

=> x=-1/8

\(\left(x-\frac{1}{4}\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow x-\frac{1}{4}=\frac{1}{8}\)

\(\Leftrightarrow x=\frac{1}{8}+\frac{1}{4}\)

\(\Leftrightarrow x=\frac{1}{8}+\frac{2}{8}\)

\(\Leftrightarrow x=\frac{3}{8}\)

Vậy \(x=\frac{3}{8}\)

ban shinicho kudo oi ban doc sai de roi a cam on

\(\left(x+\frac{1}{4}\right)^2=\frac{1}{64}\)

\(x+\frac{1}{4}=\sqrt{\frac{1}{64}}=\frac{1}{8}\)

\(x=\frac{1}{8}-\frac{1}{4}\)

\(x=-\frac{1}{8}\)

Vậy  x= -1/8

(x+1/4)²=1/64

=>(x+1/4)²=(1/8)\(^2\)

=>x+1/4=\(\orbr{\begin{cases}-\frac{1}{8}\\\frac{1}{8}\end{cases}}\)\(\orbr{\begin{cases}x=-\frac{3}{8}\\x=-\frac{1}{8}\end{cases}}\)

Học tốt nha~

cam on mn nhe Yeu :3

\(\left(x+\frac{1}{4}\right)^2=\frac{1}{64}\)

\(\left(x+\frac{1}{4}\right)^2=\left(\frac{1}{8}\right)^2\)

TH1 : \(x+\frac{1}{4}=\frac{1}{8}\)

\(x=\frac{1}{8}-\frac{1}{4}\)

\(x=-\frac{1}{8}\)

TH2 : \(x+\frac{1}{4}=-\frac{1}{8}\)

\(x=-\frac{1}{8}-\frac{1}{4}\)

\(x=-\frac{3}{8}\)

\(\left(x+\frac{1}{4}\right)^2=\frac{1}{64}\)

\(\Rightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\left(\frac{1}{8}\right)^2\\\left(x+\frac{1}{4}\right)^2=\left(\frac{-1}{8}\right)^2\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{1}{8}\\x+\frac{1}{4}=\frac{-1}{8}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-3}{8}\end{cases}}\)

Vậy \(x=\frac{-1}{8}\)và  \(x=\frac{-3}{8}\)

_Chúc bạn học tốt_

( x+1/4)^2=1/64

( x+1/4)^2=(1/8)^2 hoặc (-1/8)^2

=> x+1/4=1/8 hoặc x+1/4=-1/8

<=>x=1/8-1/4 hoặc x=(-1/8)-1/4

<=>x=-1/8 hoặc x=-3/8

Vậy x €{ -1/8 ; -3/8 }