Question

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)

Answer 1

Lớp sáu học bình phương rồi á???

Bài làm:

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\\ =>\left(x-\dfrac{1}{3}\right)^2=0+\dfrac{1}{4}=\dfrac{1}{4}\\ =>\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{2}\\x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\dfrac{1}{6};\dfrac{5}{6}\right\}\)

Answer 2

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)

\(\Rightarrow\)\(\left(x-\dfrac{1}{3}\right)^2=0+\dfrac{1}{4}\)

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

\(x-\dfrac{1}{3}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{5}{6}\)

Answer 3

(x -\(\dfrac{1}{3}\))²-\(\dfrac{1}{4}\)=0

=>(x-\(\dfrac{1}{3}\))²=0+\(\dfrac{1}{4}\)=\(\dfrac{1}{4}\)

=>[^{x-\(\dfrac{1}{3}\)}