若您覺得文章寫得不錯，請點選文章上的廣告，來支持小編，謝謝。
問題一
\[\dfrac{2x+1}{(x+1)(x-3)(x+4)} = \dfrac{A}{(x+1)}\ + \dfrac{B}{(x-3)}\ + \dfrac{C}{(x+4)}\]
上列式子中，\(A + B + C\)為多少？問題一
\[\dfrac{2x+1}{(x+1)(x-3)(x+4)} = \dfrac{A}{(x+1)}\ + \dfrac{B}{(x-3)}\ + \dfrac{C}{(x+4)}\]
解法一：
左右兩邊同乘\((x+1)(x-3)(x+4)\)得到\[2x+1 = A(x+4)(x-3)+B(x+4)(x-1)+C(x+1)(x-3)\]
\(x = -4\) 代入得 \(-7 = C \times -3 \times -7 \) ==> \(C = \dfrac{-1}{3}\)
\(x = -3\) 代入得 \(7 = B \times 7 \times 4 \) ==> \(B = \dfrac{1}{4}\)
\(x = -1\) 代入得 \(-1 = A \times 3 \times -4 \) ==> \(A = \dfrac{1}{12}\)
所以 \( A + B + C = \dfrac{1}{12} + \dfrac{1}{4} + \dfrac{-1}{3} = \boxed{0}\)
解法二：
原式改寫為 \(\dfrac{2x+1}{(x+1)(x-3)(x+4)} \\ = \dfrac{A}{x+1} + \dfrac{B}{x-3} + \dfrac{C}{x+4}\\ =\dfrac{A(x-3)(x+4) + B (x+1)(x+4) + C(x+1)(x-3)}{(x+1)(x-3)(x+4)}\)因為分母相同，所以比較分子的係數：
\(A(x^2 +x -12) + B (x^2+5x+4) + C(x^2-2x-3) = 2x+1\\ (A+B+C)x^2 + (A+5B-2C)x +(-12A+4B-3C) = 2x+1\)
可得到
\((A+B+C)x^2 = 0x^2\\ A+B+C = \boxed{0}\)
問題二
\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}\]
上式中\(A,B, C\) 為常數，求\(A\times B\times C\)為多少？\((A+B+C)x^2 = 0x^2\\ A+B+C = \boxed{0}\)
問題二
\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}\]
解法一：
\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}\]
\[\Rightarrow x^2 - 19 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)\]
\[\begin{cases} x=1\quad \quad \Rightarrow \quad 18=-6A\quad \, \, \, \Rightarrow \quad A=3 \\ x=-2\quad \, \Rightarrow \quad -15=15B\quad \Rightarrow \quad B=-1 \\ x=3\quad \quad \Rightarrow \quad -10=10C \quad \Rightarrow \quad C=-1 \end{cases}\]
\[ \Rightarrow ABC = -1 \times 3 \times -1 = \large \color{green}{\boxed{3}}\]
解法二：
\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}\]
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)}{(x-1)(x+2)(x-3)}\]
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x^2 - x -6) + B(x^2 - 4x + 3) + C(x^2 + x - 2)}{(x-1)(x+2)(x-3)}\]
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{(A+B+C)x^2 + (C-A-4B)x + (3B-6A-2C)}{(x-1)(x+2)(x-3)}\]
觀察兩邊的係數，可得：
\[\begin{cases}A+B+C=1\\C-A-4B=0\\3B-6A-2C= -19\end{cases}\]
\[\Rightarrow x^2 - 19 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)\]
\[\begin{cases} x=1\quad \quad \Rightarrow \quad 18=-6A\quad \, \, \, \Rightarrow \quad A=3 \\ x=-2\quad \, \Rightarrow \quad -15=15B\quad \Rightarrow \quad B=-1 \\ x=3\quad \quad \Rightarrow \quad -10=10C \quad \Rightarrow \quad C=-1 \end{cases}\]
\[ \Rightarrow ABC = -1 \times 3 \times -1 = \large \color{green}{\boxed{3}}\]
解法二：
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)}{(x-1)(x+2)(x-3)}\]
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{A(x^2 - x -6) + B(x^2 - 4x + 3) + C(x^2 + x - 2)}{(x-1)(x+2)(x-3)}\]
\[\dfrac{x^2 - 19}{(x-1)(x+2)(x-3)} = \dfrac{(A+B+C)x^2 + (C-A-4B)x + (3B-6A-2C)}{(x-1)(x+2)(x-3)}\]
觀察兩邊的係數，可得：
\[\begin{cases}A+B+C=1\\C-A-4B=0\\3B-6A-2C= -19\end{cases}\]
解聯立方程式可得: \(A =3, B = -1, C = -1\)，因此\(A\times B\times C = 3\times -1\times -1 = \boxed{3}\).
沒有留言:
張貼留言