Áp dụng bất đẳng thức Svacxo, ta có:
\(f\left(x\right)=\dfrac{4}{x}+\dfrac{9}{1-x}=\dfrac{2^2}{x}+\dfrac{3^2}{1-x}\ge\dfrac{\left(2+3\right)^2}{x+1-x}=25\)
Vậy \(f\left(x\right)_{min}=25\Leftrightarrow\dfrac{2}{x}=\dfrac{3}{1-x}\Leftrightarrow x=\dfrac{2}{5}\)