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