Đáp án: $A\ge (2\sqrt{5}+5)$
Giải thích các bước giải:
Ta có:
$A=\dfrac{x}{1-x}+\dfrac{5}{x}$
$\to A=\dfrac{x^2+5(1-x)}{x(1-x)}$
$\to A=\dfrac{x^2-5x+5}{x-x^2}$
$\to A-(2\sqrt{5}+5)=\dfrac{x^2-5x+5}{x-x^2}-(2\sqrt{5}+5)$
$\to A-(2\sqrt{5}+5)=\dfrac{x^2-5x+5-(2\sqrt{5}+5)(x-x^2)}{x-x^2}$
$\to A-(2\sqrt{5}+5)=\dfrac{\left(6+2\sqrt{5}\right)x^2-\left(10+2\sqrt{5}\right)x+5}{x-x^2}$
$\to A-(2\sqrt{5}+5)=\dfrac{\left(6+2\sqrt{5}\right)(x-\dfrac{5-\sqrt{5}}{4})^2}{x-x^2}$
Ta có $0<x<1\to x-x^2>0$
$\to A-(2\sqrt{5}+5)\ge 0$
$\to A\ge (2\sqrt{5}+5)$
Dấu = xảy ra khi $x=\dfrac{5-\sqrt{5}}{4}$
