`\frac{1}{2} ( x+ \frac{1}{3})^2 + \frac{1}{3}(x+ \frac{1}{3})^2 = \frac{10}{27}`
` ⇒ (x +\frac{1}{3})^2(\frac{1}{2} + \frac{1}{3} ) = \frac{10}{27}`
`⇒ (x +\frac{1}{3})^2 . \frac{5}{6} = \frac{10}{27}`
⇒ ` (x +\frac{1}{3})^2 = \frac{4}{9}`
⇒ ` (x +\frac{1}{3})^2 = ( \frac{2}{3})^2`
⇒ \(\left[ \begin{array}{l}x+ \dfrac{1}{3} = \dfrac{2}{3}\\x+ \dfrac{1}{3} = \dfrac{-2}{3}\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\)
Vậy `x ∈ { \frac{1}{3} ; -1}`
$Winner112$
