Tìm các số \(A;\,B;\,C\) để \(\frac{{2{x^2} - 3x + 12}}{{{{\left( {x + 3} \right)}^3}}} = \frac{A}{{{{\left( {x + 3} \right)}^3}}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}} + \frac{C}{{x + 3}}\)
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A.
\(A = 30;\,B = 15;\,C = - 2\)
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B.
\(A = 39;\,B = - 15;\,C = 2\)
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C.
\(A = 49;\,B = - 14;\,C = 2\)
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D.
\(A = 39;\,B = - 14;\,C = - 2\)
Tính tổng \(\frac{A}{{{{\left( {x + 3} \right)}^3}}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}} + \frac{C}{{{{\left( {x + 3} \right)}^3}}}\) sau đó đồng nhất hệ số.
\(\begin{array}{l}\frac{A}{{{{\left( {x + 3} \right)}^3}}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}} + \frac{C}{{{{\left( {x + 3} \right)}^3}}} = \frac{{A + B\left( {x + 3} \right) + C{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 3} \right)}^3}}}\\ = \frac{{A + B\left( {x + 3} \right) + C\left( {{x^2} + 6x + 9} \right)}}{{{{\left( {x + 3} \right)}^3}}} = \frac{{A + Bx + 3B + C{x^2} + 6Cx + 9C}}{{{{\left( {x + 3} \right)}^3}}}\\ = \frac{{C{x^2} + \left( {B + 6C} \right)x + \left( {A + 3B + 9C} \right)}}{{{{\left( {x + 3} \right)}^3}}}\end{array}\)
\(\frac{{2{x^2} - 3x + 12}}{{{{\left( {x + 3} \right)}^3}}} = \frac{A}{{{{\left( {x + 3} \right)}^3}}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}} + \frac{C}{{x + 3}} \Leftrightarrow \left\{ \begin{array}{l}C = 2\\B + 6C = - 3\\A + 3B + 9C = 12\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = 39\\B = - 15\\C = 2\end{array} \right.\)
Đáp án : B