Question

(x + 1/2)^2=1/6

tính 

 

Answer 1

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{6}\)

\(\left(x+\frac{1}{2}\right)=+-\sqrt{\frac{1}{6}}\)

Khi : 

\(\left(x+\frac{1}{2}\right)=\frac{1}{6}\)

\(x=\frac{1}{6}-\frac{1}{2}\)

\(x=-\frac{1}{3}\)

Khi

\(\left(x+\frac{1}{2}\right)=-\frac{1}{6}\)

\(x=-\frac{1}{6}-\frac{1}{2}\)

\(x=-\frac{2}{3}\)

Answer 2

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{6}\)

\(\Rightarrow x^2+\left(\frac{1}{2}\right)^2=\frac{1}{6}\)

\(\Rightarrow x^2+\frac{1}{4}=\frac{1}{6}\)

\(\Rightarrow x^2=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}\)

\(\Rightarrow x^2=\frac{-1}{12}\)

Answer 3

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{6}\)

=>\(x^2+\frac{1}{2}^2=\frac{1}{6}\)

\(x^2+\frac{1}{4}=\frac{1}{6}\)

\(x^2=\frac{1}{6}-\frac{1}{4}=-\frac{1}{12}\)

\(x=\sqrt{-\frac{1}{12}}\)