Question

cho x>0, tìm D_min = 9x² + 3x + 1/x + 1420

Answer 1

\(D=9x^2+3x+\frac{1}{x}+1420=9x^2-6x+1+9x+\frac{1}{x}+1419\)

\(D=\left(3x-1\right)^2+9x+\frac{1}{x}+1419\)

Áp dụng BĐT cauchy :\(9x+\frac{1}{x}\ge2\sqrt{9x.\frac{1}{x}}=6\)

\(\Rightarrow D\ge\left(3x-1\right)^2+1419+6\ge1425\)

dấu = xảy ra khi \(\left\{\begin{matrix}x=\frac{1}{3}\\9x=\frac{1}{x}\end{matrix}\right.\Leftrightarrow x=\frac{1}{3}}\)

Answer 2

min=1425 khi x=1/3 dg ban ko giai dc sr nhe :)