The two values A and B that give the partial function decomposition:
3-x/x^2-3x+2 = A/x-1 + B/x-2
I tried working this out but the two values I got are not on the multiple choice option list. Can anyone please help?
Maths Help?
2017-09-27 8:57 pm
回答 (4)
2017-09-27 9:04 pm
✔ 最佳答案
(3 - x)/(x² - 3x + 2) = [A/(x - 1)] + [B/(x - 2)](3 - x)/(x² - 3x + 2) = [A(x - 2)/(x - 1)(x - 2)] + [B(x - 1)/(x - 1)(x - 2)]
Then, 3 - x = A(x - 2) + B(x - 1)
Put x = 1: 2 = -A, then A = -2
Put x = 2: 1 = B, then B = 1
Hence, (3 - x)/(x² - 3x + 2) = [1/(x - 2)] - [2/(x - 1)]
2017-09-27 11:50 pm
(3-x)/(x^2-3x+2)=(3-x)/[(x-1)(x-2)]
Let 3-x=A(x-2)+B(x-1) identically.
Taking x=2, get
B=1
Taking x=1, get
A=-2
=>
(3-x)/(x^2-3x+2)=1/(x-2)-2/(x-1)
Let 3-x=A(x-2)+B(x-1) identically.
Taking x=2, get
B=1
Taking x=1, get
A=-2
=>
(3-x)/(x^2-3x+2)=1/(x-2)-2/(x-1)
2017-09-27 9:27 pm
= [A/(x - 1)] + [B/(x - 2)]
= [A.(x - 2) + B.(x - 1)] / [(x - 1).(x - 2)]
= [Ax - 2A + Bx - B] / (x² - 3x + 2)
= [- (2A + B) + x.(A + B)] / (x² - 3x + 2) → you compare with: (3 - x)/(x² - 3x + 2)
- (2A + B) = 3 ← equation (1)
(A + B) = - 1 ← equation (2)
You calculate (1) + (2)
- (2A + B) + (A + B) = 3 + (- 1)
- 2A - B + A + B = 3 - 1
- A = 2
→ A = - 2
Recall (2): (A + B) = - 1
B = - 1 - A → given that: A = - 2
→ B = 1
= [A.(x - 2) + B.(x - 1)] / [(x - 1).(x - 2)]
= [Ax - 2A + Bx - B] / (x² - 3x + 2)
= [- (2A + B) + x.(A + B)] / (x² - 3x + 2) → you compare with: (3 - x)/(x² - 3x + 2)
- (2A + B) = 3 ← equation (1)
(A + B) = - 1 ← equation (2)
You calculate (1) + (2)
- (2A + B) + (A + B) = 3 + (- 1)
- 2A - B + A + B = 3 - 1
- A = 2
→ A = - 2
Recall (2): (A + B) = - 1
B = - 1 - A → given that: A = - 2
→ B = 1
2017-09-27 8:59 pm
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