A=$\frac{x}{1-x}$ +$\frac{4}{x}$
=$\frac{x}{1-x}$ +$\frac{4-4x+4x}{x}$
=$\frac{x}{1-x}$ +$\frac{4-4x}{x}$ +$\frac{4x}{x}$
=$\frac{x}{1-x}$ +$\frac{4(1-x)}{x}$ +4
⇒A≥2$\sqrt{\frac{x}{1-x} .\frac{4(1-x)}{x}}$ +4 =2$\sqrt{4}$ +4 =2.2+4 =8 (BĐT Cosi)
Dấu "=" xảy ra⇔$\frac{x}{1-x}$ =$\frac{4(1-x)}{x}$
⇔$x$.$x$=4(1-$x$)(1-$x$)
⇔$x^{2}$ =4$(1-x)^{2}$
⇔$x^{2}$ -4$(1-x)^{2}$ =0
⇔$x^{2}$ -$[2(1-x)]^{2}$ =0
⇔$x^{2}$ -$(2-2x)^{2}$ =0
⇔($x$-2+2$x$)($x$+2-2$x$) =0
⇔(3$x$-2)(-$x$+2)=0
⇔\(\left[ \begin{array}{l}3x-2=0\\-x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{2}{3}(t\m)\\x=2(loại)\end{array} \right.\)
Vậy MinA=8 khi $x$=$\frac{2}{3}$
