Đáp án đúng: C
Giải chi tiết:Ta có:
\(\begin{align} & f\left( x-m \right)={{\left( x-m \right)}^{2}}+m\left( x-m \right)+m-13 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{x}^{2}}-2mx+{{m}^{2}}+mx-{{m}^{2}}+m-13 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={{x}^{2}}+mx+m-13-2mx \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=f\left( x \right)-2mx \\ \end{align}\)
\(\Rightarrow \left\{ \begin{align} & f\left( {{x}_{1}}-m \right)=f\left( {{x}_{1}} \right)-2m{{x}_{1}} \\ & f\left( {{x}_{2}}-m \right)=f\left( {{x}_{2}} \right)-2m{{x}_{2}} \\ \end{align} \right.\)
Do \({{x}_{1}};{{x}_{2}}\) là nghiệm của phương trình
\(f\left( x \right) = 0 \Rightarrow \left\{ \begin{array}{l}f\left( {{x_1}} \right) = 0\\f\left( {{x_2}} \right) = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}f\left( {{x_1} - m} \right) = - 2m{x_1}\\f\left( {{x_2} - m} \right) = - 2m{x_2}\end{array} \right.\)
Hơn nữa theo hệ thức Vi-ét ta có: \(\left\{ \begin{align} & {{x}_{1}}+{{x}_{2}}=-m \\ & {{x}_{1}}{{x}_{2}}=m-13 \\\end{align} \right.\) . Do \(m>13\Rightarrow \left\{ \begin{align} & {{x}_{1}}+{{x}_{2}}<0 \\ & {{x}_{1}}{{x}_{2}}>0 \\\end{align} \right.\Rightarrow \) Phương trình \(f\left( x \right)=0\) có hai nghiệm âm phân biệt.
\(\begin{array}{l} \Rightarrow \left| {{x_1}} \right|f\left( {{x_2} - m} \right) + \left| {{x_2}} \right|f\left( {{x_1} - m} \right) = 104\\ \Leftrightarrow - {x_1}.\left( { - 2m{x_2}} \right) - {x_2}\left( { - 2m{x_1}} \right) = 104\\ \Leftrightarrow 4m{x_1}{x_2} = 104\\ \Leftrightarrow m.\left( {m - 13} \right) = 26\\ \Leftrightarrow {m^2} - 13m - 26 = 0\\ \Leftrightarrow \left[ \begin{array}{l}m = \frac{{13 + \sqrt {273} }}{2}\,\,\,\left( {tm} \right)\\m = \frac{{13 - \sqrt {273} }}{2}\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)
Vậy \(m=\frac{13+\sqrt{273}}{2}\)
Chọn C
