Tìm GTNN của : A= -5/4x^2 + 6x B= 7/-2x^2 +6x+1 M= 2x + 5 – x^2 / x^2 28/07/2021 Bởi Clara Tìm GTNN của : A= -5/4x^2 + 6x B= 7/-2x^2 +6x+1 M= 2x + 5 – x^2 / x^2
Giải thích các bước giải: \(\begin{array}{l}A = \dfrac{{ – 5}}{{4{x^2} + 6x}} = \dfrac{{ – 5}}{{{{\left( {2x + \dfrac{3}{2}} \right)}^2} – \dfrac{9}{4}}} \ge \dfrac{{20}}{9}\\GTNN\dfrac{{20}}{9} \Leftrightarrow x = – \dfrac{3}{4}\\B = \dfrac{7}{{ – 2{x^2} + 6x + 1}} = \dfrac{{ – 7}}{{2{x^2} – 6x – 1}}\\ = \dfrac{{ – 7}}{{2{{\left( {x – \dfrac{3}{2}} \right)}^2} – \dfrac{{11}}{2}}} \ge \dfrac{{14}}{{11}}\\{B_{\min }} = \dfrac{{14}}{{11}} \Leftrightarrow x = \dfrac{3}{2}\\M = \dfrac{{ – {x^2} + 2x + 5}}{{{x^2}}}\\M + 1 = \dfrac{{2x + 5}}{{{x^2}}} = \dfrac{2}{x} + \dfrac{5}{{{x^2}}}\\ = 5\left( {\dfrac{1}{{{x^2}}} + 2\dfrac{1}{5}x + \dfrac{1}{{25}}} \right) – \dfrac{1}{{25}}\\ = 5{\left( {\dfrac{1}{x} + \dfrac{1}{5}} \right)^2} – \dfrac{1}{{25}} \ge – \dfrac{1}{{25}}\\{M_{\min }} = \dfrac{{ – 26}}{{25}} \Rightarrow x = – 5\end{array}\) Bình luận
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{ – 5}}{{4{x^2} + 6x}} = \dfrac{{ – 5}}{{{{\left( {2x + \dfrac{3}{2}} \right)}^2} – \dfrac{9}{4}}} \ge \dfrac{{20}}{9}\\
GTNN\dfrac{{20}}{9} \Leftrightarrow x = – \dfrac{3}{4}\\
B = \dfrac{7}{{ – 2{x^2} + 6x + 1}} = \dfrac{{ – 7}}{{2{x^2} – 6x – 1}}\\
= \dfrac{{ – 7}}{{2{{\left( {x – \dfrac{3}{2}} \right)}^2} – \dfrac{{11}}{2}}} \ge \dfrac{{14}}{{11}}\\
{B_{\min }} = \dfrac{{14}}{{11}} \Leftrightarrow x = \dfrac{3}{2}\\
M = \dfrac{{ – {x^2} + 2x + 5}}{{{x^2}}}\\
M + 1 = \dfrac{{2x + 5}}{{{x^2}}} = \dfrac{2}{x} + \dfrac{5}{{{x^2}}}\\
= 5\left( {\dfrac{1}{{{x^2}}} + 2\dfrac{1}{5}x + \dfrac{1}{{25}}} \right) – \dfrac{1}{{25}}\\
= 5{\left( {\dfrac{1}{x} + \dfrac{1}{5}} \right)^2} – \dfrac{1}{{25}} \ge – \dfrac{1}{{25}}\\
{M_{\min }} = \dfrac{{ – 26}}{{25}} \Rightarrow x = – 5
\end{array}\)