Đáp án:
$\min B = 9 \Leftrightarrow x = - 1 \,\,hoặc\,\,x = 3$
Giải thích các bước giải:
$\begin{array}{l}a) \quad x^4 - 2x^3 + 4x^2 - 8x\\ = (x^4 + 4x^2) - (2x^3 + 8x)\\ = x^2(x^2 + 4) - 2x(x^2 + 4)\\ = (x^2 +4)(x^2 - 2x) \quad \vdots \quad x^2 + 4\\ b)\quad B = x^4 - 4x^3 - 2x^2 + 12x + 18\\ \to B = x^4 - 4x^3 + 4x^2 - 6x^2 +12x +18\\ \to B = (x^2 - 2x)^2 - 6(x^2 -2x) + 18\\ \to B = (x^2 - 2x - 3)^2 + 9\\ \text{Ta có:}\\ (x^2 - 2x - 3)^2 \geq 0\quad \forall x\\ \to (x^2 - 2x - 3)^2 + 9 \geq 9\\ \to B \geq 9\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x^2 - 2x - 3 = 0\Leftrightarrow \left[\begin{array}{l} x = - 1\\x = 3\end{array}\right.\\ Vậy\,\,\min B = 9 \Leftrightarrow x = - 1 \,\,hoặc\,\,x = 3\end{array}$
