cho 2016 29/09/2021 Bởi Iris cho 2016 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho 2016
Đáp án: Giải thích các bước giải: \[\begin{array}{l}Dat\; x – 2016 = a \Rightarrow 2017 – x = 1 – a\\ Theo\,gt\,,\,2016 < x < 2017 \Rightarrow 0 < a < 1\\ S = \underbrace {\frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}}}_I + \underbrace {\frac{1}{{a\left( {1 - a} \right)}}}_J\\ Co:\,I = \frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}} \ge \frac{1}{2}{\left( {\frac{1}{a} + \frac{1}{{1 - a}}} \right)^2} \ge \frac{1}{2}{\left( {\frac{4}{{a + 1 - a}}} \right)^2} = \frac{1}{2}.16 = 8\\ J = \frac{1}{{a\left( {a - 1} \right)}} = \frac{1}{{ - {a^2} + a - \frac{1}{4} + \frac{1}{4}}} = \frac{1}{{\frac{1}{4} - {{\left( {a - \frac{1}{2}} \right)}^2}}} \ge \frac{1}{{\frac{1}{4}}} = 4\\ Vay\,S \ge 8 + 4 = 12\\ Dau\, = \,xay\,ra\,khi\,a = 1 - a \Leftrightarrow a = \frac{1}{2} \end{array}\] Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}Dat\;
x – 2016 = a \Rightarrow 2017 – x = 1 – a\\
Theo\,gt\,,\,2016 < x < 2017 \Rightarrow 0 < a < 1\\ S = \underbrace {\frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}}}_I + \underbrace {\frac{1}{{a\left( {1 - a} \right)}}}_J\\ Co:\,I = \frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}} \ge \frac{1}{2}{\left( {\frac{1}{a} + \frac{1}{{1 - a}}} \right)^2} \ge \frac{1}{2}{\left( {\frac{4}{{a + 1 - a}}} \right)^2} = \frac{1}{2}.16 = 8\\ J = \frac{1}{{a\left( {a - 1} \right)}} = \frac{1}{{ - {a^2} + a - \frac{1}{4} + \frac{1}{4}}} = \frac{1}{{\frac{1}{4} - {{\left( {a - \frac{1}{2}} \right)}^2}}} \ge \frac{1}{{\frac{1}{4}}} = 4\\ Vay\,S \ge 8 + 4 = 12\\ Dau\, = \,xay\,ra\,khi\,a = 1 - a \Leftrightarrow a = \frac{1}{2} \end{array}\]