\(R\left(x\right)=-x^2+\dfrac{2}{3}x+\dfrac{1}{5}\)
\(R\left(x\right)=-1\left(x^2-\dfrac{2}{3}x-\dfrac{1}{5}\right)\)
\(R\left(x\right)=-1\left(x^2-2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^2-\dfrac{1}{5}\right)\)
\(R\left(x\right)=-1\left[\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{45}\right]\)
\(R\left(x\right)=-1\left(x-\dfrac{1}{3}\right)^2+\dfrac{14}{45}\)
\(R\left(x\right)=\dfrac{14}{45}-\left(x-\dfrac{1}{3}\right)^2\le\dfrac{14}{45}\)
Vậy R(x) max = 14/45 tại x = 1/3
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