[artty60]

ข้อ2 อีกวิธี

$A=\frac{[(x+\frac{1}{x})^3]^2-(x^3+\frac{1}{x^3})^2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$

$=\frac{[(x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})][(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}]}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$

$=3(x+\frac{1}{x})$

$Amin$เมื่อ$x=1$ , $\,x\in \mathbf{R^+}$

$\therefore Amin=6$