Đáp án:

$B.\, -\dfrac{289}{24}$

Giải thích các bước giải:

$\begin{array}{l}\quad C = (2x - 3)(4 + 3x)\\ \to C = 8x +6x^2 - 12 - 9x\\ \to C = 6x^2 - x - 12\\ \to C = 6\left(x^2 - 2\cdot \dfrac{1}{12}\cdot x +\dfrac{1}{144} -\dfrac{289}{144}\right)\\ \to C = 6\left(x^2 - 2\cdot \dfrac{1}{12}\cdot x +\dfrac{1}{144} \right) - \dfrac{289}{24}\\ \to C = 6\cdot\left(x - \dfrac{1}{12}\right)^2 - \dfrac{289}{24}\\ \text{Ta có:}\\ \quad 6\cdot\left(x - \dfrac{1}{12}\right)^2 \geq 0\quad \forall x\\ \to 6\cdot\left(x - \dfrac{1}{12}\right)^2 - \dfrac{289}{24} \geq -\dfrac{289}{24}\\ \to C \geq -\dfrac{289}{24}\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x - \dfrac{1}{12} = 0 \Leftrightarrow x = \dfrac{1}{12}\\ Vậy\,\,\min C = -\dfrac{289}{24} \Leftrightarrow x= \dfrac{1}{12} \quad \end{array}$

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