主頁 高登熱話 吃喝玩樂 科技消費 名人專訪 短片 現有會員可[按此]登入。未成為會員可[按此]註冊。 [公司模式 - 關]  [懷舊模式 - 開]  [Youtube 預覽 - 關]  [大字型]  [小字型] 您現在聚腳在 學術台內。 高登公告 有關帖文題目顯示香港字問題 一直以來，不少會員反映討論區帖文題目未能正常顯示香港字，由於此項更新牽涉整個系統，故一直未能完善。然而我們明白各會員對此有一定需求，決定於星期三早上7時短暫「熄登」更新系統，解決題目顯示問題，預計需時兩小時。不便之處，敬請原諒。 更新完成後題目將支援香港字、大部分日文及韓文字體，字數限制將由現時25個全形字符增加至30個。與此同時，討論區桌面版將增設M版現有的回帶及追蹤功能，發表頁面的題目輸入位置亦會加入字數提示功能，希望能改善大家的使用體驗。 香港高登討論區管理員團隊 精選文章 DeerGamer   Find the number of ways that 11 identical balls can put into five different boxes. 12/1/2020 15:14  Find the number of ways that 11 identical balls can put into five different boxes. 12/1/2020 15:15  Find the number of ways that 11 identical balls can put into five different boxes. A box may contain zero balls. 12/1/2020 15:17  睇到呢抽數字睇到暈 12/1/2020 17:25  Ax^2+Bx+C+D=0
Ax^2+Bx+B=0
So B/A=a+b and ab
So a^4+b^4=(a^2+b^2)^2-2(ab)^2=((B/A)^2-2(B/A))^2-2(B/A)^2
=(B/A)^4-4(B/A)^3+2(B/A)^2

12/1/2020 17:31  x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z. 12/1/2020 17:55  為左推動學術台發展, 我決定答岩有一膠幣以示鼓勵 12/1/2020 18:04  Find the number of ways that 11 identical balls can put into five different boxes. A box may contain zero balls.

I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.

12/1/2020 19:15  Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y. f(x) = 1/xf(y) = 1/yf(xy) = 1/xy = f(x)f(y)搵到一個答案 應該唔得，因為0嘅時候係undefined不過如果幫0加exception嘅話，咁就無問題無錯, 因為f(0)應該undefined.應該係話你define唔到f(0)使得上述所有條件成立 12/1/2020 19:30  x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.If x and y are both 2, then z=3 would have satisfied the requirement. (Being greater than 2*sqrt(2))If x=1 and y=4 or y=1 and x=4z must be greater than 3 but less than 5that is 4But triangle with two sides of length 4 and the remaining side of length 1 is not an obtuse angled triangle. 12/1/2020 19:37  Find the number of ways that 11 identical balls can put into five different boxes. A box may contain zero balls.

I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.

I can write a formula without any steps. Think about it yourself.

12/1/2020 19:56  x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.If x and y are both 2, then z=3 would have satisfied the requirement. (Being greater than 2*sqrt(2))If x=1 and y=4 or y=1 and x=4z must be greater than 3 but less than 5that is 4But triangle with two sides of length 4 and the remaining side of length 1 is not an obtuse angled triangle.好 12/1/2020 19:57  Find the number of ways that 11 identical balls can put into five different boxes.15C4=1365 12/1/2020 20:05 Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y. let y = 0,
f(x) . f(0) = f(x . 0)
f(x) . f(0) = f(0)
f(x) . f(0) / f(0) = f(0) / f(0)
f(x) = 1

12/1/2020 20:19  Find all positive integers n such that (2^8+2^11+2^n) is a perfect square. 12/1/2020 20:19  Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.2^8+2^11+2^n=2^8+2(2)^4(2)^6+2^nIf n=12, =2^8+2(2)^4(2)^6+2^12=(2^4+2^6)^2=80^2...uniqueness 有d難prove, 你等等Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.2^8+2^11+2^n=2^8+2(2)^4(2)^6+2^nIf n=12, =2^8+2(2)^4(2)^6+2^12=(2^4+2^6)^2=80^2...uniqueness 有d難prove, 你等等。 12/1/2020 21:22  Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.2^8+2^11+2^n=2^8+2(2)^4(2)^6+2^nIf n=12,=2^8+2(2)^4(2)^6+2^12=(2^4+2^6)^2=80^2...uniqueness 有d難prove, 你等等。 12/1/2020 21:22  I had proven when n=12, 2^8+2^11+2^n is a perfect square.
I won't consider n<12 as it's easy to compute.
Now for n>12.

12/1/2020 21:40  Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.2^8+2^11+2^n=2^8(1+8+2^(n-8))=2^8(9+2^(n-8))n=12 is a solution, since 9+2^4=25 which is a perfect square.To claim n=12 is the only solution, let n-8=Nand we require (3^2+2^N) to be perfect square, say a^2then 2^N=(a+3)(a-3), which follows (a+3)=2^N1, (a-3)=2^N2implying 2^N1 is 6 greater than 2^N2But the powers of 2 are 2, 4, 8, 16, 32, 64, 128, ...and it can been seen that N1=3, N2=1 is the only possible combinationthus proving uniqueness. 12/1/2020 21:44  Find the number of ways that 11 identical balls can put into five different boxes. A box may contain zero balls.

I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.

13/1/2020 11:27  Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.2^8+2^11+2^n=2^8(1+8+2^(n-8))=2^8(9+2^(n-8))n=12 is a solution, since 9+2^4=25 which is a perfect square.To claim n=12 is the only solution, let n-8=Nand we require (3^2+2^N) to be perfect square, say a^2then 2^N=(a+3)(a-3), which follows (a+3)=2^N1, (a-3)=2^N2implying 2^N1 is 6 greater than 2^N2But the powers of 2 are 2, 4, 8, 16, 32, 64, 128, ...and it can been seen that N1=3, N2=1 is the only possible combinationthus proving uniqueness.其實最後部份因為6=2×3所以直接寫2^N2=2及2^(N1-N2)-1=3便足夠 13/1/2020 11:35  Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.

13/1/2020 12:59  Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.目前搵到一條:f(x)=x^3-2x^2+2x 13/1/2020 14:43  Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.目前搵到一條:f(x)=x^3-2x^2+2xStep: f(0)=0^3f(1)-2(0)^4+2(0)^2=0f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2f(1)=0f(1)=(1)f(2)-2+2=f(2)hence x, (x-1), (x-2) are factors of f(x)This means degree of f(x) >=3Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*degree of L.H.S. of (*)= 2ndegree of R.H.S. of (*)= n+3L.H.S.=R.H.S. implies 2n=n+3, n=3Since x, (x-1), (x-2) are three factors of f(x) (proven),Let f(x)=Cx(x-1)(x-2)= C(x^3-2x^2+2x) where C is constantBy(*),C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2Hence -3C=-2-C and 2C=2 gives C=1,f(x)=x^3-2x^2+2x Q.E.D. 13/1/2020 16:06  Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.目前搵到一條:f(x)=x^3-2x^2+2xStep: f(0)=0^3f(1)-2(0)^4+2(0)^2=0f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2f(1)=0f(1)=(1)f(2)-2+2=f(2)hence x, (x-1), (x-2) are factors of f(x)This means degree of f(x) >=3Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*degree of L.H.S. of (*)= 2ndegree of R.H.S. of (*)= n+3L.H.S.=R.H.S. implies 2n=n+3, n=3Since x, (x-1), (x-2) are three factors of f(x) (proven),Let f(x)=Cx(x-1)(x-2)= C(x^3-3x^2+2x) where C is constantBy(*),C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2Hence -3C=-2-C and 2C=2 gives C=1,f(x)=x^3-3x^2+2x Q.E.D. 13/1/2020 16:11

快速回覆 - 輸入以下項目