cho pt:x ² – (2m+1)*x + 2m – 4 = 0 tìm gtnn

Đáp án:

$\begin{array}{l}
{x^2} – \left( {2m + 1} \right)x + 2m – 4 = 0\\
 \Rightarrow {x^2} – 2.\left( {m + \frac{1}{2}} \right).x + {\left( {m + \frac{1}{2}} \right)^2} – {\left( {m + \frac{1}{2}} \right)^2} + 2m – 4 = 0\\
 \Rightarrow {\left( {x – m – \frac{1}{2}} \right)^2} – {m^2} – m – \frac{1}{4} + 2m – 4 = 0\\
 \Rightarrow {\left( {x – m – \frac{1}{2}} \right)^2} – {m^2} + m – \frac{{17}}{4} = 0\\
Do:{\left( {x – m – \frac{1}{2}} \right)^2} \ge 0\forall x,m\\
 \Rightarrow {\left( {x – m – \frac{1}{2}} \right)^2} – {m^2} + m – \frac{{17}}{4} \ge  – {m^2} + m – \frac{{17}}{4}\\
 \Rightarrow GTNN: – {m^2} + m – \frac{{17}}{4}
\end{array}$