高一指數函數

令 y=f(x)=g(x), u=2^x+2^(-x)=> g(x)=y=k/u => u=k/y=> f(x)=y=4x+u/3=4x+k/3y => 3y^2-12x*y-k=0二次方程式解: y=[6x+-√(36x^2+3k)]/3=(6x+-D)/3 => A=(x,(6x+D)/3), B=(x,(6x-D)/3)=> AB=2*D/3=2√(36x^2+3k)]/3=4√17 => 36x^2+3k=36*17=> k=12(17-x^2)
y=f(x)=4x+u/3=g(x)=k/u=12(17-x^2)/u2次方程式: u^2+12x*u+36(x^2-17)=02次方程式解: u=6(√17-x)指數方程式: u=2^x+2^(-x)=6(√17-x)令h(x)=2^x+2^(-x)+6(x-√17)解出: x=2.876482 => h(2.876482)=0=> k=12(17-x^2)=104.71024.......ans

2014-03-29 20:11:36 補充：
題目更新:

令 y=f(x)=g(x), u=2^x+2^(-x)

u/3=k/u => 3k=u^2 => u=+-√(3k)

3k=(2^x+2^(-x))^2=4^x+2+4^(-x)

0=4^2x+(2-3k)4^x+1

4^x=[(3k-2)+-D]/2......D=√(9k^2-12k)


Ax={log[(3k-2)+D]-log(2)}/log(4)

Bx={log[(3k-2)-D]-log(2)}/log(4)



Ay=4*Ax+-√(3k)/3

By=4*By+-√(3k)/3

2014-03-29 20:12:26 補充：
Ax-Bx=log[(3k-2+D)/(3k-2-D)]/log(4)

Ay-By=4(Ax-Bx)=4*log[(3k-2+D)/(3k-2-D)]/log(4)


AB^2=(Ax-Bx)^2+(Ay-By)^2

=17*{log[(3k-2+D)/(3k-2-D)]/log(4)}^2

=(4√17)^2


log[(3k-2+D)/(3k-2-D)]/log(4)=4

log[(3k-2+D)/(3k-2-D)]=4*log(4)

(3k-2+D)/(3k-2-D)=4^4=256

=> 9216k^2-12288k-260100=0

2014-03-29 20:12:44 補充：
k=6.0208333, -4.6875......ans

2014-03-29 20:27:25 補充：
3k=u^2>0 => k=-4.6875拋棄

=> k=6.02083333

2014-03-30 07:23:56 補充：
補充: 合分比定律a/b=c/d => (a+b)/(a-b)=(c+d)(c-d)

(3k-2+D)/(3k-2-D)=256 => (3k-2)/D=257/255

255^2*(9k^2-12k)=255^2*(3k-2)^2

0=2304k^2-2*1536k-65025

k=6.0208333.......ans

k=-4.6875負值拋棄

2014-03-31 09:59:59 補充：
請問Ay = 4*±√3k/3是怎麼來的?

Ans:

Ay=y

=f(x)

=4x+u/3

=4*Ax+√(3k)/3.......u=√(3k)

=4*Ax+√(3k)/3

參考下面的網址看看

http://phi008780520.pixnet.net/blog

非常感謝月下大師. 您的算式較簡潔.

用對稱的觀念.....

2014-03-31 00:25:57 補充：
4x + (2^x + 2-^x)/3
4x+ k/ ( 2^x + 2-^x )
兩式相減
(2^x + 2-^x)/3 - k/ ( 2^x + 2-^x )=0 ,有兩根 Ax ,Bx
Ax 與 Bx 必對稱 y軸 令 Ax=-Bx=t >0 代入 f(x)
得：
Ay=4t + (2^t + 2-^t)/3
By=-4t+ (2^t + 2-^t)/3
Ay -By=8t , Ax-Bx=2t

2014-03-31 00:26:04 補充：
(Ax-Bx)^2 +(Ay-By)^2 =4t^2+64t^2=68t^2 =(4根號17)^2= 4*4*17
t^2 = 4 ,t=2 ,( 2^x + 2-^x )=4+1/4=17/4
(2^x + 2-^x)/3 - k/ ( 2^x + 2-^x )=0
17/12 - k/(17/4)=0
k=289/48