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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 10: cho T= $\frac{a^2+a+2}{a^2}$ tìm a để Tmin

Toán Lớp 10: cho T= $\frac{a^2+a+2}{a^2}$ tìm a để Tmin

Comments ( 1 )

  1. Giải đáp: $a =  – 4$
     
    Lời giải và giải thích chi tiết:
    $\begin{array}{l}
    T = \dfrac{{{a^2} + a + 2}}{{{a^2}}}\\
     = 1 + \dfrac{1}{a} + \dfrac{2}{{{a^2}}}\\
     = \dfrac{2}{{{a^2}}} + \dfrac{1}{a} + 1\\
     = 2.\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{2}.\dfrac{1}{a} + \dfrac{1}{{16}}} \right) – 2.\dfrac{1}{{16}} + 1\\
     = 2.\left( {\dfrac{1}{{{a^2}}} + 2.\dfrac{1}{a}.\dfrac{1}{4} + \dfrac{1}{{16}}} \right) + \dfrac{7}{8}\\
     = 2.{\left( {\dfrac{1}{a} + \dfrac{1}{4}} \right)^2} + \dfrac{7}{8}\\
    Do:{\left( {\dfrac{1}{a} + \dfrac{1}{4}} \right)^2} \ge 0\\
     \Leftrightarrow 2.{\left( {\dfrac{1}{a} + \dfrac{1}{4}} \right)^2} \ge 0\\
     \Leftrightarrow 2.{\left( {\dfrac{1}{a} + \dfrac{1}{4}} \right)^2} + \dfrac{7}{8} \ge \dfrac{7}{8}\\
     \Leftrightarrow T \ge \dfrac{7}{8}\\
     \Leftrightarrow GTNN:T = \dfrac{7}{8}khi:\dfrac{1}{a} + \dfrac{1}{4} = 0 \Leftrightarrow a =  – 4\\
    Vậy\,a =  – 4
    \end{array}$

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222-9+11+12:2*14+14 = ? ( )