$\left\{\begin{matrix}\sqrt{x^2-2x+6}.log_{3}(6-y)=x & & \\\sqrt{y^2-2y+6}.log_{3}(6-z)=y & & \\\sqrt{z^2-2z+6}.log_{3}(6-x)=z \end{matrix}\right.$
#1
Posted 18-10-2012 - 20:35
$\left\{\begin{matrix}\sqrt{x^2-2x+6}.log_{3}(6-y)=x
& & \\\sqrt{y^2-2y+6}.log_{3}(6-z)=y
& & \\\sqrt{z^2-2z+6}.log_{3}(6-x)=z
\end{matrix}\right.$
#2
Posted 18-10-2012 - 20:45
Giải hệ pt sau:
$\left\{\begin{matrix}\sqrt{x^2-2x+6}.log_{3}(6-y)=x
& & \\\sqrt{y^2-2y+6}.log_{3}(6-z)=y
& & \\\sqrt{z^2-2z+6}.log_{3}(6-x)=z
\end{matrix}\right.$
Hướng dẫn:
Đưa hệ đã cho về hệ hoán vị vòng quanh:
\[\left\{ \begin{array}{l}
\frac{x}{{\sqrt {{x^2} - 2x + 6} }} = {\log _3}\left( {6 - y} \right)\\
\frac{y}{{\sqrt {{y^2} - 2y + 6} }} = {\log _3}\left( {6 - z} \right)\\
\frac{z}{{\sqrt {{z^2} - 2z + 6} }} = {\log _3}\left( {6 - x} \right)
\end{array} \right.\]
Xét các hàm số: $f\left( t \right) = \frac{t}{{\sqrt {{t^2} - 2t + 6} }},\,\,g\left( t \right) = {\log _3}\left( {6 - t} \right)$.
Chứng minh hệ có nghiệm duy nhất: $x = y = z=?$
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#3
Posted 18-10-2012 - 22:00
#4
Posted 19-10-2012 - 01:43
Anh ơi nếu xét các hàm kia thì xét cho nó đông biến trên cac khoảng xác định rồi cho nó co nghiệm duy nhất ạ
Bạn xét các hàm trên xem đơn điệu tăng hay giảm trên các khoảng xác định (nếu có). Sau đó dựa vào tính chất hoán vị vòng quanh để kết luận.
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#5
Posted 21-10-2012 - 16:56
Đây chính là bài VMO 2006 mà bạn. Bạn chịu khó xem đáp án tiếng Anh nhé, khi nào rảnh mình dịch (chính xác hơn là nhờ người dịch =))~).Giải hệ pt sau:
$\left\{\begin{matrix}\sqrt{x^2-2x+6}.log_{3}(6-y)=x
& & \\\sqrt{y^2-2y+6}.log_{3}(6-z)=y
& & \\\sqrt{z^2-2z+6}.log_{3}(6-x)=z
\end{matrix}\right.$
----
Solution:
The domain of definition of the system is $x$, $y$, $z<6$. Then the system is equivalent to:
$$\left\{\begin{matrix} \dfrac{x}{\sqrt{x^{2}-2x+6}}=\log_{3}\left ( 6-y \right )
& & \\ \dfrac{y}{\sqrt{y^{2}-2y+6}}=\log_{3}\left ( 6-z \right )& & \\ \dfrac{z}{\sqrt{z^{2}-2z+6}}=\log_{3}\left ( 6-x \right )
\end{matrix}\right.$$
Consider a function $f\left ( x \right )=\dfrac{x}{\sqrt{x^{2}-2x+6}}, \ \ x<6$ which has a derivative $f'\left ( x \right )=\dfrac{6-x}{\left ( x^{2}-2x+6 \right )\sqrt{x^{2}-2x+6}}>0, \ \ \forall x<6$ and so $f\left ( x \right )$ is increasing, while a function $g\left ( x \right )=\log_{3}\left ( 6-x \right ), \ \ \forall x<6$ is obviously decreasing.
Let $\left ( x,y,z \right )$ is a solution of system. We prove that $x=y=z$.
Without loss of generality, we can assume that $x=\max \left \{ x,y,z \right \}$. There are two cases:
Case 1: $x\geq y\geq z$. In this case, since $f\left ( x \right )$ increases, $\log_{3}\left ( 6-y \right )\geq \log_{3}\left ( 6-z \right )\geq \log_{3}\left ( 6-x \right )$ and hence , since $g\left ( x \right )$ decreases, $x\geq y\geq z$. Then $y\geq z$ and $z\geq y$ give $y=z$ and therefore $x=y=z$.
Case 2: $x\geq z\geq y$. Similarly, we get $x\geq z$ and $z\geq x$ which give $x=z$ and therefore $x=y=z$.
Thus the system becomes $f\left ( x \right )=g\left ( x \right )=6, \ \ x<6$. Note that $f\left ( x \right )$ icreases, and $g\left ( x \right )$ decreases, then the equation $f\left ( x \right )=g\left ( x \right )$ has at most one solution. Since $x=3$, as can be easily seen, is a solution, the unique solution of the equation, and therefore, of the system, is $\left ( 3,3,3 \right )$. $\blacksquare$Edited by Gin Escaper, 21-10-2012 - 16:57.
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Thích ngủ.
#6
Posted 21-10-2012 - 18:11
Cái này dịch dễ màĐây chính là bài VMO 2006 mà bạn. Bạn chịu khó xem đáp án tiếng Anh nhé, khi nào rảnh mình dịch (chính xác hơn là nhờ người dịch =))~).
----
Solution:
The domain of definition of the system is $x$, $y$, $z<6$. Then the system is equivalent to:
$$\left\{\begin{matrix} \dfrac{x}{\sqrt{x^{2}-2x+6}}=\log_{3}\left ( 6-y \right )
& & \\ \dfrac{y}{\sqrt{y^{2}-2y+6}}=\log_{3}\left ( 6-z \right )& & \\ \dfrac{z}{\sqrt{z^{2}-2z+6}}=\log_{3}\left ( 6-x \right )
\end{matrix}\right.$$
Consider a function $f\left ( x \right )=\dfrac{x}{\sqrt{x^{2}-2x+6}}, \ \ x<6$ which has a derivative $f'\left ( x \right )=\dfrac{6-x}{\left ( x^{2}-2x+6 \right )\sqrt{x^{2}-2x+6}}>0, \ \ \forall x<6$ and so $f\left ( x \right )$ is increasing, while a function $g\left ( x \right )=\log_{3}\left ( 6-x \right ), \ \ \forall x<6$ is obviously decreasing.
Let $\left ( x,y,z \right )$ is a solution of system. We prove that $x=y=z$.
Without loss of generality, we can assume that $x=\max \left \{ x,y,z \right \}$. There are two cases:Case 1: $x\geq y\geq z$. In this case, since $f\left ( x \right )$ increases, $\log_{3}\left ( 6-y \right )\geq \log_{3}\left ( 6-z \right )\geq \log_{3}\left ( 6-x \right )$ and hence , since $g\left ( x \right )$ decreases, $x\geq y\geq z$. Then $y\geq z$ and $z\geq y$ give $y=z$ and therefore $x=y=z$.
Case 2: $x\geq z\geq y$. Similarly, we get $x\geq z$ and $z\geq x$ which give $x=z$ and therefore $x=y=z$.
Thus the system becomes $f\left ( x \right )=g\left ( x \right )=6, \ \ x<6$. Note that $f\left ( x \right )$ icreases, and $g\left ( x \right )$ decreases, then the equation $f\left ( x \right )=g\left ( x \right )$ has at most one solution. Since $x=3$, as can be easily seen, is a solution, the unique solution of the equation, and therefore, of the system, is $\left ( 3,3,3 \right )$. $\blacksquare$
Điều kiện $x,y,z<6$. Hệ đã cho trở thành:
$$\left\{\begin{matrix} \dfrac{x}{\sqrt{x^{2}-2x+6}}=\log_{3}\left ( 6-y \right )
& & \\ \dfrac{y}{\sqrt{y^{2}-2y+6}}=\log_{3}\left ( 6-z \right )& & \\ \dfrac{z}{\sqrt{z^{2}-2z+6}}=\log_{3}\left ( 6-x \right )
\end{matrix}\right.$$
Ta có: $f\left ( x \right )=\dfrac{x}{\sqrt{x^{2}-2x+6}}, \ \ x<6$
Đạo hàm $f'\left ( x \right )=\dfrac{6-x}{\left ( x^{2}-2x+6 \right )\sqrt{x^{2}-2x+6}}>0, \ \ \forall x<6$ vì vậy $f\left ( x \right )$ tăng, trong khi $g\left ( x \right )=\log_{3}\left ( 6-x \right ), \ \ \forall x<6$ là hàm giảm
$\left ( x,y,z \right )$ là nghiệm của hệ. Ta chứng minh$x=y=z$.
Không mất tính tổng quát giả sử $x=\max \left \{ x,y,z \right \}$. Xét 2 trường hợp sau
TH1: $x\geq y\geq z$. Ta có $f\left ( x \right )$ tăng $\log_{3}\left ( 6-y \right )\geq \log_{3}\left ( 6-z \right )\geq \log_{3}\left ( 6-x \right )$, $g\left ( x \right )$ giảm, $x\geq y\geq z$. $y\geq z$ và $z\geq y$ thì $y=z$ và vì vậy $x=y=z$.
TH2: $x\geq z\geq y$. Tương tự ta có $x\geq z$ ; $z\geq x$ thì $x=z$ do đó $x=y=z$.
Vì vậy $f\left ( x \right )=g\left ( x \right )=6, \ \ x<6$. Ở đây $f\left ( x \right )$ tăng, và $g\left ( x \right )$ giảm, thì $f\left ( x \right )=g\left ( x \right )$ có một nghiệm. Ta có $x=3$, là nghiệm của phương trình vậy phương trình có nghiệm $\left ( 3,3,3 \right )$. $\blacksquare$Edited by Ispectorgadget, 21-10-2012 - 18:13.
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