Tim GTNN cua a)3x^2-6x+17/x^2-2x+5 b)2x^2-16x+41/x^2-8x+22

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\frac{{2{x^2} – 16x + 41}}{{{x^2} – 8x + 22}} = \frac{{\left( {2{x^2} – 16x + 44} \right) – 3}}{{{x^2} – 8x + 22}} = \frac{{2\left( {{x^2} – 8x + 22} \right) – 3}}{{{x^2} – 8x + 22}}\\
 = 2 – \frac{3}{{{x^2} – 8x + 22}} = 2 – \frac{3}{{{{\left( {x – 4} \right)}^2} + 6}}\\
{\left( {x – 4} \right)^2} \ge 0,\forall x \Rightarrow \,\,\,\,{\left( {x – 4} \right)^2} + 6 \ge 6,\forall x\\
 \Rightarrow \frac{{ – 3}}{{{{\left( {x – 4} \right)}^2} + 6}} \ge \frac{{ – 3}}{6} = \frac{{ – 1}}{2}\\
 \Rightarrow \frac{{2{x^2} – 16x + 41}}{{{x^2} – 8x + 22}} \ge 2 – \frac{1}{2} = \frac{3}{2}
\end{array}\)

Dấu ‘=’ xảy ra khi và chỉ khi x=4