Bài dễ rối lắm, ai mò chỗ sai hộ tui nhá ( vì thấy đáp số kì quá )
$A=(x+a)(x+b)(x+c)(x+d)$
$=[x^2+(a+b)x+ab][x^2+(c+d)x+cd]$
$=[x^2+(a+b)x+ab][x^2+(a+b)x+cd]$
Đặt $y=x^2+(a+b)x+ab$, ta có
$A=y(y-ab+cd)$
$=y^2-y(ab-cd)$
$=y^2-y(ab-cd)+\frac{(ab-cd)^2}{4}-\frac{(ab-cd)^2}{4}$
$=(y-\frac{ab-cd}{2})^2-\frac{(ab-cd)^2}{4} \geq -\frac{(ab-cd)^2}{4} \forall y$
$\Rightarrow min A = -\frac{(ab-cd)^2}{4}$
$\Leftrightarrow y-\frac{ab-cd}{2}=0$
$\Leftrightarrow x^2+(a+b)x+ab-\frac{ab-cd}{2}=0$
$\Leftrightarrow x^2+(a+b)x+\frac{ab+cd}{2}=0$
$\Leftrightarrow x^2+(a+b)x+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+\frac{2ab+2ac}{4}=0$
$\Leftrightarrow (x+\frac{a+b}{2})^2-\frac{(a+b)^2-2ab-2ac}{4}=0$
$\Leftrightarrow (x+\frac{a+b}{2})^2-\frac{a^2+2ab+b^2-2ab-2ac}{4}=0$
$\Leftrightarrow (x+\frac{a+b}{2})^2-\frac{a^2+b^2-2ac}{4}=0
$\Leftrightarrow (x+\frac{a+b+\sqrt{a^2+b^2-2ac}}{2})(x+\frac{a+b-\sqrt{a^2+b^2-2ac}}{2})=0$
Vậy, $minA = -\frac{(ab-cd)^2}{4} \Leftrightarrow x=-\frac{a+b+\sqrt{a^2+b^2-2ac}}{2} or x=\frac{\sqrt{a^2+b^2-2ac}-a-b}{2}$
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