Question

Tìm x biết: (x-\(\dfrac{2}{3}\))\(^2\)=\(\dfrac{5}{6}\)

Answer 1

(\(x\)-\(\dfrac{2}{3}\))^{2\(=\)\(\dfrac{5}{6}\)}

^\(x\)-\(\dfrac{4}{9}\)\(=\)\(\dfrac{5}{6}\)

\(x\)\(=\)\(\dfrac{5}{6}\)+\(\dfrac{4}{9}\)

\(x\)\(=\)\(\dfrac{23}{18}\)

Answer 2

\(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}\)

\(< =>\left(x-\dfrac{2}{3}\right)^2-\dfrac{5}{6}=0\)

<=>\(\left(x-\dfrac{2}{3}\right)^2-\left(\dfrac{\sqrt{5}}{\sqrt{6}}\right)^2=0\)

\(< =>\left(x-\dfrac{2}{3}+\dfrac{\sqrt{5}}{\sqrt{6}}\right)\left(x-\dfrac{2}{3}-\dfrac{\sqrt{5}}{\sqrt{6}}\right)=0\)

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

<=>\(\left[{}\begin{matrix}x=\dfrac{4-\sqrt{30}}{6}\\x=\dfrac{4+\sqrt{30}}{6}\end{matrix}\right.\)

VẬy..