Thẻ: Hoc tap 2022

Chọn A Theo giả thiết ta có: \(f\left( x \right)-g\left( x \right)\)\( ={{\left( x-1 \right)}^{2}}\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\)\( ={{x}^{4}}+b{{x}^{2}}-mx+n\,\,\left( * \right)\) Ta có: \(\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\frac{f\left( x \right)-g\left( x \right)}{{{\left( x-1 \right)}^{2}}}dx}\)\(=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)dx}\)\(=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( x-{{x}_{1}}+{{x}_{1}}-{{x}_{2}} \right)dx}\) \(\begin{align} & =\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left[ {{\left( x-{{x}_{1}} \right)}^{2}}+\left( x-{{x}_{1}} \right)\left( {{x}_{1}}-{{x}_{2}} \right) \right]dx}\\&=\left. \left( \frac{{{\left( x-{{x}_{1}} \right)}^{3}}}{3}+\left( {{x}_{1}}-{{x}_{2}} \right)\frac{{{\left( x-{{x}_{1}} […]