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)

Đáp án:

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