Câu hỏi 20

Ta có: \(f'\left( x \right)=\frac{1}{\sqrt{2x-1}}\Rightarrow f''\left( x \right)=\frac{-\left( \sqrt{2x-1} \right)'}{2x-1}=-\frac{1}{\sqrt{{{\left( 2x-1 \right)}^{3}}}}\)

\(\Rightarrow f'''\left( x \right)=\frac{\left( \sqrt{{{\left( 2x-1 \right)}^{3}}} \right)'}{{{\left( 2x-1 \right)}^{2}}}=\frac{3\sqrt{2x-1}}{{{\left( 2x-1 \right)}^{3}}}\Rightarrow f\left( 1 \right)=3\)

Chọn A.