Chủ đề này có 5 trả lời

[kiokiung]

tim tat ca cac gia tri ma tong x+y+z co the dat duoc voi x,y,z la nghiem cua he
x=y*(4-y)
y=z*(4-z)
z=x*(4-x)

[PTH_Thái Hà]

$\left\{ \begin{array}{l} \\ x = y\left( {4 - y} \right) \\ y = z\left( {4 - z} \right) \\ z = x\left( {4 - x} \right) \\ \end{array} \right. $

[dark templar]

Đây là đề thi đề nghị Olympic 30-4
Từ hệ suy ra $3S=x^2+y^2+z^2 \geq 0 \Rightarrow x \geq 0$ hoặc $y \geq 0$ hoặc $z \geq 0$
Giả sử $x \geq 0$
$ y(4-y)=x \Rightarrow y(4-y) \geq 0 $
$\Rightarrow 0 \leq y \le 4 $
$\Rightarrow z(4-z)=y \geq 0$
$ \Rightarrow x.(4-x)=z \geq 0$
Vậy $0 \leq x,y,z \leq 4$
Đặt $x = 4\sin ^2 \theta (0 \le \theta \le \dfrac{\pi }{2})$
$ \Rightarrow z = x.(4 - x) = 4\sin ^2 \theta .4\cos ^2 \theta = 4\sin ^2 2\theta $
$\Rightarrow y = z(4 - z) = 4\sin ^2 2\theta .4\cos ^2 2\theta = 4\sin ^2 4\theta $
$\Rightarrow x = y(4 - y) = 4\sin ^2 4\theta .4\cos ^2 4\theta = 4\sin ^2 8\theta $
$\Rightarrow 4\sin ^2 \theta = 4\sin ^2 8\theta $
$\Leftrightarrow \cos 16\theta = \cos 2\theta $
$\Leftrightarrow 16\theta = 2\theta + k2\pi $ hoặc $16\theta = - 2\theta + k2\pi (k \in Z) $
$ \Leftrightarrow \theta = \dfrac{{k\pi }}{7}$ hoặc $\theta = \dfrac{{k\pi }}{9}(k \in Z)$
Nếu $\theta = \dfrac{{k\pi }}{7}$
Do $0 \le \theta \le \dfrac{\pi }{2} $
$\Rightarrow 0 \le \dfrac{k}{7} \le \dfrac{1}{2} \Leftrightarrow 0 \le k \le \dfrac{7}{2} \Leftrightarrow k \in \left\{ {0,1,2,3} \right\}(k \in Z)$
$*k=0 \Rightarrow S=0$
$*k=1 $
$\Rightarrow S=4\left( {\sin ^2 \dfrac{\pi }{7} + \sin ^2 \dfrac{{2\pi }}{7} + \sin ^2 \dfrac{{4\pi }}{7}} \right) $
$= 4\left[ {\dfrac{3}{2} - \dfrac{3}{2}\left( {\cos \dfrac{{2\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{8\pi }}{7}} \right)} \right] $

Bài viết đã được chỉnh sửa nội dung bởi dark templar: 29-10-2010 - 20:27

[dark templar]

(Tiếp tục)
Hơn nữa :
$P=\cos \dfrac{{2\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{8\pi }}{7} $
$\Rightarrow 2\sin \dfrac{{2\pi }}{7}P = \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7} - \sin \dfrac{{2\pi }}{7} + \sin \dfrac{{10\pi }}{7} - \sin \dfrac{{6\pi }}{7} $
$= \sin \dfrac{{10\pi }}{7} + \sin \dfrac{{4\pi }}{7} - \sin \dfrac{{2\pi }}{7} $
$= 2\sin \pi \cos ^{} \dfrac{{3\pi }}{7} - \sin \dfrac{{2\pi }}{7}$
$= - \sin \dfrac{{2\pi }}{7} $
$\Rightarrow P = \dfrac{{ - 1}}{2} $
$\Rightarrow S = 4\left( {\dfrac{3}{2} + \dfrac{3}{4}} \right) = 9 $
$\bullet k = 2 \Rightarrow S = 4\left( {\sin ^2 \dfrac{{2\pi }}{7} + \sin ^2 \dfrac{{4\pi }}{7} + \sin ^2 \dfrac{{8\pi }}{7}} \right)$
$= 4\left( {\sin ^2 \dfrac{\pi }{7} + \sin ^2 \dfrac{{2\pi }}{7} + \sin ^2 \dfrac{{4\pi }}{7}} \right) = 9$
$\bullet k = 3 \Rightarrow S = 4\left( {\sin ^2 \dfrac{{3\pi }}{7} + \sin ^2 \dfrac{{6\pi }}{7} + \sin ^2 \dfrac{{12\pi }}{7}} \right)$
$= 4\left( {\sin ^2 \dfrac{\pi }{7} + \sin ^2 \dfrac{{2\pi }}{7} + \sin ^2 \dfrac{{4\pi }}{7}} \right) = 9 $
P/s:Bạn chịu khó đọc từng đoạn nhé vì bài này dài quá!

Bài viết đã được chỉnh sửa nội dung bởi dark templar: 29-10-2010 - 22:12

[dark templar]

(Tiếp tục)
Nếu $\theta =\dfrac{k\pi }{9}$
$0 \le \theta \le \dfrac{\pi }{2} \Rightarrow 0 \le \dfrac{k}{9} \le \dfrac{1}{2} $
$\Leftrightarrow 0 \le k \le \dfrac{9}{2} $
$\Leftrightarrow k \in \left\{ {0,1,2,3,4} \right\}(k \in Z) $
$\bullet k = 0 \Rightarrow S = 0 $
$\bullet k = 1 \Rightarrow S = 4\left( {\sin ^2 \dfrac{\pi }{9} + \sin ^2 \dfrac{{2\pi }}{9} + \sin ^2 \dfrac{{4\pi }}{9}} \right) $
$= 4\left[ {\dfrac{3}{2} - \dfrac{1}{2}\left( {\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9}} \right)} \right] $
$Q = \cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9} $
$\Rightarrow 2\sin \dfrac{{2\pi }}{9}Q = \sin \dfrac{{4\pi }}{9} + \sin \dfrac{{6\pi }}{9} - \sin \dfrac{{2\pi }}{9} + \sin \dfrac{{10\pi }}{9} - \sin \dfrac{{6\pi }}{9} $
$= \sin \dfrac{{10\pi }}{9} + \sin \dfrac{{4\pi }}{9} - \sin \dfrac{{2\pi }}{9} $
$= 2\sin \dfrac{{7\pi }}{9}\cos \dfrac{{3\pi }}{9} - \sin \dfrac{{2\pi }}{9} $
$= \sin \dfrac{{7\pi }}{9} - \sin \dfrac{{2\pi }}{9} = 0 $
$\Rightarrow Q = 0 $
$\Rightarrow S = 4.\dfrac{3}{2} = 6 $

Bài viết đã được chỉnh sửa nội dung bởi dark templar: 29-10-2010 - 20:26

[dark templar]

(Đến đây là hết rồi !)
$\bullet k = 2 $
$\Rightarrow S = 4\left( {\sin ^2 \dfrac{{2\pi }}{9} + \sin ^2 \dfrac{{4\pi }}{9} + \sin ^2 \dfrac{{8\pi }}{9}} \right) $
$= 4\left( {\sin ^2 \dfrac{\pi }{9} + \sin ^2 \dfrac{{2\pi }}{9} + \sin ^2 \dfrac{{4\pi }}{9}} \right) = 6 $
$\bullet k = 3 $
$\Rightarrow S = 4\left( {\sin ^2 \dfrac{{3\pi }}{9} + \sin ^2 \dfrac{{6\pi }}{9} + \sin ^2 \dfrac{{12\pi }}{9}} \right)$
$= 4\left( {\sin ^2 \dfrac{\pi }{3} + \sin ^2 \dfrac{{2\pi }}{3} + \sin ^2 \dfrac{{4\pi }}{3}} \right) $
$= 4.3\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2 = 9 $
$\bullet k = 4 $
$\Rightarrow S = 4\left( {\sin ^2 \dfrac{{4\pi }}{9} + \sin ^2 \dfrac{{8\pi }}{9} + \sin ^2 \dfrac{{16\pi }}{9}} \right) $
$= 4\left( {\sin ^2 \dfrac{\pi }{9} + \sin ^2 \dfrac{{2\pi }}{9} + \sin ^2 \dfrac{{4\pi }}{9}} \right) = 6 $
$\Rightarrow S \in \left\{ {0,6,9} \right\}$