(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~
\(\left(x+\frac{1}{4}\right)^2=\frac{1}{64}\)
\(\left(x+\frac{1}{4}\right)^2=\left(\frac{1}{8}\right)^2\)
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 }