求解!兩題數學的解法?

2019-08-16 5:05 pm
1.多項式f(x)除以(x-1)^2及(x-2)^2的餘式依次為3x及3x+2,則f(x)除以(x-1)^2(x-2)的餘式為?
2.設二次多項式f(x)=a(x-1)(x-2)+b(x-1)+c
且g(x)=3*(x-2)(x-3)/(1-2)(1-3)
+2*(x-3)(x-1)/(2-3)(2-1)
-4*(x-1)(x-2)/(3-1)(3-2)
若已知f(√2)=g(√2)+2,f(√3)=g(√3)+3,f(√6)=g(√6)+6,則a、b、c各等於多少?

回答 (1)

2019-08-16 6:13 pm
✔ 最佳答案
1.
設 f(x) = Q(x) (x - 1)²(x - 2) + k(x - 1)² + 3x = P(x) (x - 2)² + 3x + 2 
則 f(2) = k + 3(2) = 3(2) + 2 ⇒ k = 2 , 
故 f(x) 除以(x - 1)² (x - 2) 的餘式為 2(x - 1)² + 3x = 2x² - x + 2。

2.
考慮 f(x) = g(x) + x²,若它為x之二次方程,由題目條件它竟然有三根√2,√3及√6,與二次方程只有二根矛盾!
故 f(x) = g(x) + x² 必為恆等式無疑,則
f(3) = 2a+2b+c = g(3)+3² = -4+9 = 5,
f(2) = b+c = g(2)+2² = 2+4 = 6,
f(1) = c = g(1)+1² = 3+1 = 4,
得 c=4,b=6-4=2,a=(5-2(2)-(4))/2=-3/2。


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