tìm MIN A=2/(6x-5-9x^2) tìm MIN B=(3X^2-8X+6)/(x^2-2x+1) tìm MIN c=x(X-3)(x-4)(x-7)

0 bình luận về “tìm MIN A=2/(6x-5-9x^2) tìm MIN B=(3X^2-8X+6)/(x^2-2x+1) tìm MIN c=x(X-3)(x-4)(x-7)”

1. Đáp án:

   $\begin{array}{l}
   a)A = \frac{2}{{6x – 5 – 9{x^2}}}\\
   Do:6x – 5 – 9{x^2}\\
    =  – 9{x^2} + 6x – 1 – 4\\
    =  – \left( {9{x^2} – 6x + 1} \right) – 4\\
    =  – {\left( {3x – 1} \right)^2} – 4\\
   Do: – {\left( {3x – 1} \right)^2} \le 0\forall x\\
    \Rightarrow  – {\left( {3x – 1} \right)^2} – 4 \le  – 4\forall x\\
    \Rightarrow \frac{2}{{6x – 5 – 9{x^2}}} \ge \frac{2}{{ – 4}} =  – \frac{1}{2}\\
    \Rightarrow A \ge  – \frac{1}{2}\forall x\\
    \Rightarrow \min A =  – \frac{1}{2} \Leftrightarrow z = \frac{1}{3}\\
   b)B = \frac{{3{x^2} – 8x + 6}}{{{x^2} – 2x + 1}}\\
    = \frac{{3{x^2} – 6x + 3 – 2x + 2 + 1}}{{{{\left( {x – 1} \right)}^2}}}\\
    = \frac{{3\left( {{x^2} – 2x + 1} \right)}}{{{x^2} – 2x + 1}} – \frac{{2\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} + \frac{1}{{{{\left( {x – 1} \right)}^2}}}\\
    = 3 – \frac{2}{{x – 1}} + \frac{1}{{{{\left( {x – 1} \right)}^2}}}\\
    = \frac{1}{{{{\left( {x – 1} \right)}^2}}} – 2.\frac{1}{{x – 1}} + 1 + 2\\
    = {\left( {\frac{1}{{x – 1}} – 1} \right)^2} + 2 \ge 2\forall x \ne 1\\
    \Rightarrow Min\,B = 2 \Leftrightarrow \frac{1}{{x – 1}} = 1 \Rightarrow x – 1 = 1 \Rightarrow x = 2\left( {tm} \right)\\
   c)\\
   C = x\left( {x – 3} \right)\left( {x – 4} \right)\left( {x – 7} \right)\\
    = x\left( {x – 7} \right)\left( {x – 3} \right)\left( {x – 4} \right)\\
    = \left( {{x^2} – 7x} \right)\left( {{x^2} – 7x + 12} \right)\\
    = {\left( {{x^2} – 7x} \right)^2} + 12\left( {{x^2} – 7x} \right) + 36 – 36\\
    = {\left( {{x^2} – 7x + 6} \right)^2} – 36 \ge  – 36\forall x\\
    \Rightarrow \min C =  – 36\, \Leftrightarrow {x^2} – 7x + 6 = 0\\
    \Rightarrow x = 1/x = 6
   \end{array}$

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