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Câu hỏi: Cho $\int{x{{(2x-3)}^{5}}}dx=A{{(2x-3)}^{7}}+B{{(2x-3)}^{6}}+C,$ với $A,B,C\in \mathbb{R}$.Tính giá trị biểu
thức $7A-2B$
A. 0.
B. $\dfrac{1}{2}$.
C. 3.
D. 5.
thức $7A-2B$
A. 0.
B. $\dfrac{1}{2}$.
C. 3.
D. 5.
Ta có: $\int{x{{\left( 2x-3 \right)}^{5}}dx=\dfrac{1}{2}\int{\left[ {{\left( 2x-3 \right)}^{6}}+3{{\left( 2x-3 \right)}^{5}} \right]dx}}$
$=\dfrac{1}{2}\int{{{\left( 2x-3 \right)}^{6}}dx+\dfrac{3}{2}\int{{{\left( 2x-3 \right)}^{5}}dx}}=\dfrac{1}{28}{{\left( 2x-3 \right)}^{7}}+\dfrac{1}{8}{{\left( 2x-3 \right)}^{6}}+C$
$\Rightarrow $ $A=\dfrac{1}{28} ; B=\dfrac{1}{8}\Rightarrow 7A-2B=\dfrac{7}{28}-\dfrac{2}{8}=0$
$=\dfrac{1}{2}\int{{{\left( 2x-3 \right)}^{6}}dx+\dfrac{3}{2}\int{{{\left( 2x-3 \right)}^{5}}dx}}=\dfrac{1}{28}{{\left( 2x-3 \right)}^{7}}+\dfrac{1}{8}{{\left( 2x-3 \right)}^{6}}+C$
$\Rightarrow $ $A=\dfrac{1}{28} ; B=\dfrac{1}{8}\Rightarrow 7A-2B=\dfrac{7}{28}-\dfrac{2}{8}=0$
Đáp án A.