13-03-2023 | 608
Cho \(\frac{{3{x^2} + 3x + 5}}{{{x^3} - 3x + 2}} = \frac{A}{{{{\left( {x - 1} \right)}^2}}} + \frac{B}{{x - 1}} + \frac{C}{{x + 2}}\). Khi đó S = A - B - C bằng:
Bài giải:
Ta có: \(\frac{{3{x^2} + 3x + 5}}{{{x^3} - 3x + 2}} = \frac{{3{x^2} + 3x + 5}}{{{{(x - 1)}^2}(x + 2)}}\)
\(\frac{A}{{{{(x - 1)}^2}}} + \frac{B}{{x - 1}} + \frac{C}{{x + 2}}\)
\(= \frac{{A(x + 2) + B(x - 1)(x + 2) + C{{(x - 1)}^2}}}{{{{(x - 1)}^2}(x + 2)}}\)
\(= \frac{{Ax + 2A + B{x^2} + Bx - 2B + C{x^2} - 2Cx + C}}{{{{(x - 1)}^2}(x + 2)}}\)
\(= \frac{{(B + C){x^2} + (A + B - 2C)x + 2A - 2B + C}}{{{{(x - 1)}^2}(x + 2)}}\)
Đồng nhất tử thức, ta suy ra:
\(\left\{ \begin{gathered} B + C = 3 \hfill \\ A + B - 2C = 3 \hfill \\ 2A - 2B + C = 5 \hfill \\ \end{gathered} \right.\)
\( \Leftrightarrow \left\{ \begin{gathered} A = \frac{{11}}{3} \hfill \\ B = \frac{{16}}{9} \hfill \\ C = \frac{{11}}{9} \hfill \\ \end{gathered} \right.\)
\(\Rightarrow A - B - C = \frac{2}{3}\)
