Ta có: \(A - \frac{1}{2} = \frac{{{x^2} - 8x + 25}}{{{x^2} - 6x + 25}} - \frac{1}{2}\)

⇔ \(A - \frac{1}{2} = \frac{{2\left( {{x^2} - 8x + 25} \right)}}{{2\left( {{x^2} - 6x + 25} \right)}} - \frac{{{x^2} - 8x + 25}}{{2\left( {{x^2} - 6x + 25} \right)}}\)

⇔ \(A - \frac{1}{2} = \frac{{2{x^2} - 16x + 50 - {x^2} + 6x - 25}}{{2\left( {{x^2} - 6x + 25} \right)}}\)

⇔ \(A - \frac{1}{2} = \frac{{{x^2} - 10x + 25}}{{2\left( {{x^2} - 6x + 25} \right)}}\)

⇒ \(A - \frac{1}{2} = \frac{{{{\left( {x - 5} \right)}^2}}}{{2\left( {{x^2} - 6x + 25} \right)}} \ge 0,\forall x\)

⇒ \(A \ge \frac{1}{2}\)

Dấu “=” xảy ra khi x – 5 = 0 hay x = 5

Lại có: A = \(\frac{{{x^2} - 8x + 25}}{{{x^2} - 6x + 25}}\)

⇒\(A - \frac{9}{8} = \frac{{{x^2} - 8x + 25}}{{{x^2} - 6x + 25}} - \frac{9}{8}\)

⇔ \[A - \frac{9}{8} = \frac{{8\left( {{x^2} - 8x + 25} \right)}}{{8\left( {{x^2} - 6x + 25} \right)}} - \frac{{9\left( {{x^2} - 8x + 25} \right)}}{{8\left( {{x^2} - 6x + 25} \right)}}\]

⇔ \[A - \frac{9}{8} = \frac{{8{x^2} - 64x + 200 - 9{x^2} + 54x - 225}}{{8\left( {{x^2} - 6x + 25} \right)}}\]

⇔ \[A - \frac{9}{8} = \frac{{ - {x^2} - 10x - 25}}{{8\left( {{x^2} - 6x + 25} \right)}}\]

⇔ \[A - \frac{9}{8} = \frac{{ - {{\left( {x + 5} \right)}^2}}}{{8\left( {{x^2} - 6x + 25} \right)}} \le 0,\forall x\]

Suy ra: \(A \le \frac{9}{8}\)

Dấu “=” xảy ra khi x + 5 = 0 hay x = –5

Vậy x = 5 thì biểu thức có GTNN là \(\frac{1}{2}\) và x = –5 thì biểu thức có GTLN là \(\frac{9}{8}\).