Với 0 < x < 1 ; ta có : \(A=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=\left(\sqrt{3}+2\right)^2\)
" = " \(\Leftrightarrow\dfrac{\sqrt{3}}{1-x}=\dfrac{2}{x}\) \(\Leftrightarrow x=\dfrac{2}{\sqrt{3}+2}=4-2\sqrt{3}\) (t/m)
- Với \(0< x< 1\), ta có:
\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)
\(=\dfrac{3+3x-3x}{1-x}+\dfrac{4-4x+4x}{x}\)
\(=\dfrac{3\left(1-x\right)+3x}{1-x}+\dfrac{4\left(1-x\right)+4x}{x}\)
\(=3+\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+4\)
\(=7+\left[\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}\right]\)
\(\ge7+2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}\)
\(=7+4\sqrt{3}\)
- Dấu "=" xảy ra khi \(\sqrt{\dfrac{3x}{1-x}}=\sqrt{\dfrac{4\left(1-x\right)}{x}}\Leftrightarrow x=4-2\sqrt{3}\left(tm\right)\)
- Vậy \(MinA=7+4\sqrt{3}\), đạt tại \(x=4-2\sqrt{3}\).
