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數學 高中

想請問第12.題!我的解法完全不是答案想要的,想請問這樣解錯誤在哪裡呢?我先用有的條件假數出f(x)再假設乘了(x^2+1)的情況,餘式太高再除一次這樣。非常感謝🙏

N.已知實係數多項式f(x)的次數不超過兩次 k為正實數,又/(3) = 3k,則(9)= 52k fiyy=k f(x)=k 6 #f11)=K + L=K f(x)= a(x-1)(x-¹) + b (x-1) + © (13)=3× ax=(x-1)× "fly=k- K b+k=k ⇒b=0 a (X-1) (X-1) + K = (1/ ₁a+k=3K =) 2931 a=X 92 皆為化 【中科實中模擬考】 12. 設f(x)為三次實係數多項式,若滿足(1) = 3,f(2)=10,f(-1)=-17,f(4)=78求 【全國聯合模擬考】 (x²+1)f(x)除以x−3的餘式為 13. 實係數多項式 f(x)滿足f(1)=1,f(2) = 4,f(3)=9,f(4)=16,則下列敘述哪些是正 【中科實中模擬考】 確的? (A)f(x)除以x-1的餘式為1 (B)對於任意實數x,都有f(x)=x² (C)f(x)除以(x-2)(x-3)的餘式為x² (D)令g(x)=f(x)-x²,則(x-1)(x-2)(x-3)(x-4)為g(x)的因式 (E)f(x)不可能為一個三次多項式 ²² f(x) = a (x-1)(x-2)(x+1) + b(x-1)/(x-1) + ((x-1) + d + f(x) => x-1)(x-2)(x+1) - (x-1)(x-2) + 2(x-1) + 3₂ " £13) = Y 19 12 maid (C1) Boitanidoo (H) → f (x) = (x-3) (8 (X) + $1 (X) (3) 31 STUJOUTE of anotiziv bowolls roid ali of zomerb tupitib taom odiod of ti - D D G Y borobianos (O) Y mesoniis ei sbivong won 29tole sousinovnos asol vise to asgafonę gritoolios bus gaige X (X²+1) 7 (x²+1) f(x) = (x-3) (@(x)) (X²+1) + 3) (X²+1) 9M sau o exat ode doume dailgad bodil oda ydw bodes anW 542 gnir alanade stirw sonsbitnos may vod boqlad on golondos Si bus agnining to e boutest noindirxo 1-331+0+ 21 agaitnisq odi dow liet t'abib3/93ibliod od oti bodesto tovinb aas bus loodbe 93 +3ed nodw emexo diem list of boen nummos si sxeuparus ada of bamwulst (8) boyonteab snow 26 bs seorfw slqooq 101 totoda yoqats not bolned (4) **** 43x+31) # no bobooqsb (3)

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英文 高中

想請問第❻題 我理解的是 以「我們現在的角度」覺得「當時他吃的蘋果」,用現在式 是不是我對這個文法的用法觀念錯誤或題意該怎麼理解?非常感謝!!🙏🏼

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已解決 回答數: 1