Tìm Min
1. C= $\frac{1}{-x^2+4}$
D= $\frac{-1}{x^2-4}$
2. E= $\frac{4}{-x^2+2x-3}$
F= $\frac{-4}{x^2-2x+3}$
3. H= $\frac{3}{-x^2-4x+1}$
Tìm Min
1. C= $\frac{1}{-x^2+4}$
D= $\frac{-1}{x^2-4}$
2. E= $\frac{4}{-x^2+2x-3}$
F= $\frac{-4}{x^2-2x+3}$
3. H= $\frac{3}{-x^2-4x+1}$
Đáp án:
$\begin{array}{l}
1)Do:{x^2} \ge 0\\
\Rightarrow – {x^2} \le 0\\
\Rightarrow – {x^2} + 4 \le 4\\
\Rightarrow \dfrac{1}{{ – {x^2} + 4}} \ge \dfrac{1}{4}\\
\Rightarrow C \ge \dfrac{1}{4}\\
\Rightarrow \min C = \dfrac{1}{4}\\
D = – \dfrac{1}{{{x^2} – 4}}\\
{x^2} – 4 \ge – 4\\
\Rightarrow \dfrac{1}{{{x^2} – 4}} \le – \dfrac{1}{4}\\
\Rightarrow – \dfrac{1}{{{x^2} – 4}} \ge \dfrac{1}{4}\\
\Rightarrow D \ge \dfrac{1}{4}\\
\Rightarrow MinD = \dfrac{1}{4}\\
2) – {x^2} + 2x – 3\\
= – {x^2} + 2x – 1 – 2\\
= – \left( {{x^2} – 2x + 1} \right) – 2\\
= – {\left( {x – 1} \right)^2} – 2 \le – 2\\
\Rightarrow \dfrac{1}{{ – {x^2} + 2x – 3}} \ge – \dfrac{1}{2}\\
\Rightarrow \dfrac{4}{{ – {x^2} + 2x – 3}} \ge – 2\\
\Rightarrow \min E = – 2\\
F = \dfrac{{ – 4}}{{{x^2} – 2x + 3}}\\
{x^2} – 2x + 3 = {\left( {x – 1} \right)^2} + 2 \ge 2\\
\Rightarrow \dfrac{1}{{{x^2} – 2x + 3}} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{{ – 4}}{{{x^2} – 2x + 3}} \ge – 4\\
\Rightarrow F \ge – 4\\
\Rightarrow \min \,F = – 4\\
3)H = \dfrac{3}{{ – {x^2} – 4x + 1}}\\
– {x^2} – 4x + 1\\
= – {x^2} – 4x – 4 + 5\\
= – \left( {{x^2} + 4x + 4} \right) + 5\\
= – {\left( {x + 2} \right)^2} + 5 \le 5\\
\Rightarrow \dfrac{3}{{ – {x^2} – 4x + 1}} \ge \dfrac{3}{5}\\
\Rightarrow \min H = \dfrac{3}{5}
\end{array}$