Đáp án:
$\min\left[\dfrac{11z^2 - 22z + 33}{\left(\dfrac{1}{2^2} - 1\right)\left(\dfrac{1}{3^2} - 1\right)\cdots\left(\dfrac{1}{10^2} - 1\right)}\right] = 40\Leftrightarrow z = 1$
Giải thích các bước giải:
Ta có:
$\left(\dfrac{1}{2^2} - 1\right)\left(\dfrac{1}{3^2} - 1\right)\cdots\left(\dfrac{1}{10^2} - 1\right)$
$= \left(\dfrac{1}{2} - 1\right)\left(\dfrac{1}{3} - 1\right)\cdots\left(\dfrac{1}{10} - 1\right)\left(\dfrac{1}{2} +1\right)\left(\dfrac{1}{3} + 1\right)\cdots\left(\dfrac{1}{10} + 1\right)$
$=\left(-\dfrac12\right)\left(-\dfrac23\right)\cdots\left(-\dfrac{9}{10}\right)\cdot\dfrac32\cdot\dfrac43\cdots\dfrac{11}{10}$
$=\dfrac{1}{10}\cdot\dfrac{11}{2}$
$=\dfrac{11}{20}$
Do đó:
$\dfrac{11z^2 - 22z + 33}{\left(\dfrac{1}{2^2} - 1\right)\left(\dfrac{1}{3^2} - 1\right)\cdots\left(\dfrac{1}{10^2} - 1\right)}$
$=\dfrac{11(z^2 - 2z +3)}{\dfrac{11}{20}}$
$=20(z^2 - 2z + 3)$
$= 20(z^2 - 2z + 1) + 40$
$= 20(z -1)^2 + 40$
Do $(z-1)^2 \geq 0\quad \forall z$
nên $20(z -1)^2 + 40\geq 40$
Dấu $=$ xảy ra $\Leftrightarrow z- 1 = 0\Leftrightarrow z = 1$
Vậy $\min\left[\dfrac{11z^2 - 22z + 33}{\left(\dfrac{1}{2^2} - 1\right)\left(\dfrac{1}{3^2} - 1\right)\cdots\left(\dfrac{1}{10^2} - 1\right)}\right] = 40\Leftrightarrow z = 1$
