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[問題]解X

發表於 : 2004-04-08 2:27 AM
moon2003
X^2+(X/(X+1))^2=3
解X

發表於 : 2004-04-13 3:43 PM
大地旅法師
大概算了一下
是-2+2^(1/2)
和-2-2^(1/2)
嗎..?

發表於 : 2004-04-14 3:59 AM
moon2003
代進去以後不對喔
大地旅法師 寫:-2+2^(1/2)
=8-4√2=2.34314575...
大地旅法師 寫:-2-2^(1/2)
=8+4√2=13.6568543...

又:請解題者註明計算過程,thanx^^

發表於 : 2004-04-15 11:19 PM
lckene
X^2+(X/(X+1))^2=3
=>(0-X)^2+(1-1/(X+1))^2=3
=>√((0-X)^2+(1-1/(X+1))^2)=√3
也就是說把題目改成:求一點(X,1/X+1)到(0,1)的距離為√3
=>求一點在Y=(1/(X+1))上到(0,1)之距為√3
=>Y=(1/(X+1))
解聯立方程式: (X-0)^2+(Y-1)^2 = 3
與 XY+Y=1
=>(X-0)^2+(Y-1)^2 = X^2+Y^2-2Y-2=0
又X=1/Y-1代入上式
得Y^2-2Y-1-2/Y+1/(Y^2)=0
=>Y^2+1/(Y^2) - 2(Y+1/Y) - 1 = 0
=>(Y+1/Y)^2-2(Y)(1/Y) - 2(Y+1/Y) - 1 = 0
=>(Y+1/Y)^2 - 2(Y+1/Y) - 3 = 0
=>(Y+1/Y) = 3 or -1
=>Y^2-3Y+1=0 or Y^2+Y+1=0
又Y^2+Y+1=0之解為虛根
=>Y=(3+(√5))/2
代回XY+Y=1
得X=(1-√5)/2 or (1+√5)/2





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好累........

發表於 : 2004-04-16 2:42 AM
moon2003
嗯嗯,lckene兄答對囉!
那麼,我把我的算法也貼上來吧: (雖然也是很麻煩啦,呵呵)
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令x=√3cosθ,x/(x+1)=√3sinθ
偷懶一下:cosθ=c,sinθ=s
x=x√3s+√3s
x-x√3s=x(1-√3s)=√3s
x=√3s/(1-√3s)=√3c
√3s=√3c-3sc
√3s-√3c=√3(s-c)=-3sc
令t=s-c,則sc=(1-t^2)/2
√3t=-3(1-t^2)/2
解之得t=√3 or -√3/3
但-√2<t<√2
故t=-√3/3
-√3/3=s-c
s=c-√3/3
s^2=c^2-2√3c/3+1/3=1-c^2
解之得c=(√3±√15)/6
所以x=√3c=(3±√45)/6=(1±√5)/2
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就這樣,希望有人能想到更漂亮的算法喲^^