\(A=6\left(x^2-\dfrac{13}{6}x+1\right)\\ =6\left(x^2-2.\dfrac{13}{12}+\dfrac{169}{144}\right)-\dfrac{25}{144}\\ =6\left(x-\dfrac{13}{12}\right)^2-\dfrac{25}{144}\\ 6\left(x-\dfrac{13}{12}\right)^2-\dfrac{25}{144}\ge-\dfrac{25}{144}\\ \)
Dấu = xảy ra khi
\(x-\dfrac{13}{12}=0\\ x=\dfrac{13}{12}\\ \)
Vậy \(Min_A=-\dfrac{25}{144}khix=\dfrac{13}{12}\)
\(=6\left(x^2-\dfrac{13}{6}x+1\right)\)
\(=6\left(x^2-2\cdot x\cdot\dfrac{13}{12}+\dfrac{169}{144}-\dfrac{25}{144}\right)\)
\(=6\left(x-\dfrac{13}{12}\right)^2-\dfrac{25}{24}>=-\dfrac{25}{24}\)
Dấu '=' xảy ra khi x=13/12
