2 câu trả lời
Đáp án:
Giải thích các bước giải: \[\begin{array}{l}Dat\; x - 2016 = a \Rightarrow 2017 - x = 1 - a\\ Theo\,gt\,,\,2016 < x < 2017 \Rightarrow 0 < a < 1\\ S = \underbrace {\frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}}}_I + \underbrace {\frac{1}{{a\left( {1 - a} \right)}}}_J\\ Co:\,I = \frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}} \ge \frac{1}{2}{\left( {\frac{1}{a} + \frac{1}{{1 - a}}} \right)^2} \ge \frac{1}{2}{\left( {\frac{4}{{a + 1 - a}}} \right)^2} = \frac{1}{2}.16 = 8\\ J = \frac{1}{{a\left( {a - 1} \right)}} = \frac{1}{{ - {a^2} + a - \frac{1}{4} + \frac{1}{4}}} = \frac{1}{{\frac{1}{4} - {{\left( {a - \frac{1}{2}} \right)}^2}}} \ge \frac{1}{{\frac{1}{4}}} = 4\\ Vay\,S \ge 8 + 4 = 12\\ Dau\, = \,xay\,ra\,khi\,a = 1 - a \Leftrightarrow a = \frac{1}{2} \end{array}\]
\[\begin{array}{l}
cho\,\,x - 2016 = a \Rightarrow 2017 - x = 1 - a\\
gt:\,2016 < x < 2017 \Rightarrow 0 < a < 1\\
S = \underbrace {\frac{1}{{{a^2}}} + \frac{1}{{\left( {1 - {a^2}} \right)}}\,}_I + \underbrace {\frac{1}{{a(1 - a)}}}_J\\
I = \frac{1}{{{a^2}}} + \frac{1}{{1 - {a^2}}} + \frac{1}{{a\left( {1 - a} \right)}} \ge \frac{1}{2}{\left( {\frac{1}{a} + \frac{1}{{1 - a}}} \right)^2} \ge \frac{1}{2}\left( {\frac{4}{{a + 1 - a}}} \right) = \frac{1}{2}.16 = 18\\
S \ge 8 + 4 = 12\\
'' = ''xay\,ra\,\,khia = 1 - a \Leftrightarrow a = 12
\end{array}\]