Đáp án:
$\rm A=x^2-3x+5+\dfrac{4}{x}\\=x^2-4x+4+x+\dfrac{4}{x}+1\\=(x-2)^2+x-2.\sqrt{x}.\sqrt{\dfrac{4}{x}}+\dfrac{4}{x}+1+4\\=(x-2)^2+\Big(\sqrt{x}-\sqrt{\dfrac{2}{x}}\Big)^2+5\\(x-2)^2+\Big(\sqrt{x}-\sqrt{\dfrac{2}{x}}\Big)^2 \geq 0\\\to (x-2)^2+\Big(\sqrt{x}-\sqrt{\dfrac{2}{x}}\Big)^2+5 \geq 5\\Hay\,\,A \geq 5\\\text{Dấu "=" xảy ra khi}\\\begin{cases}x-2=0\\\sqrt{x}-\sqrt{\dfrac{2}{x}}\end{cases}\\↔x=2\\Vậy\,\,Min-A=5↔x=2$
