Chủ đề này có 6 trả lời

[my_ha_123]

giai hệ phương trình:
$ \[\left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2yz = x} \\
{{y^2} + 2zx = y} \\
{{z^2} + 2xy = z} \\
\end{array}} \right.\] $

Bài viết đã được chỉnh sửa nội dung bởi my_ha_123: 21-11-2010 - 13:45

[khacduongpro_165]

{${x^2} + 2yz = x$}
{{$y^2} + 2zx = y$}
{{$z^2} + 2xy = z$}
Đề thế ah?

Bài viết đã được chỉnh sửa nội dung bởi khacduongpro_165: 21-11-2010 - 15:25

[NguyThang khtn]

{${x^2} + 2yz = x$}
{{$y^2} + 2zx = y$}
{{$z^2} + 2xy = z$}
Đề thế ah?

neu de la the thi :
cong tung ve cua ba PT lai ta duoc:
$(x+y+z)^2 = x+y+z$
den day de roi!

[my_ha_123]

neu de la the thi :
cong tung ve cua ba PT lai ta duoc:
$(x+y+z)^2 = x+y+z$
den day de roi!

em giai toi do lai khong giai duoc!giup em lam tiep voi

[NguyThang khtn]

xet hai truong hop :
1)x+y+z= 0 roi thay vao phuong trinh da cho!
2) tuong tu!

[my_ha_123]

xet hai truong hop :
1)x+y+z= 0 roi thay vao phuong trinh da cho!
2) tuong tu!

thay sao anh?thay 1 truong hop jiup em voi

[dark templar]

Bài này cũng dễ ,chỉ rắc rối ở chỗ xét nhiều trường hợp mà thôi!!!!!!
$\left\{ \begin{array}{l}x^2 + 2yz = x\left( 1 \right) \\ y^2 + 2zx = y\left( 2 \right) \\ z^2 + 2xy = z\left( 3 \right) \\ \end{array} \right. $
$\left( 1 \right) - \left( 2 \right) x^2 - y^2 + 2z\left( {y - x} \right) = x - y \Leftrightarrow \left( {x - y} \right)\left( {x + y - 2z - 1} \right) = 0 $
$\Leftrightarrow \left[ \begin{array}{l}x = y \\ x + y = 2z + 1 \\ \end{array} \right. $
$\bullet x = y:$
$\left( 1 \right) \Leftrightarrow x^2 + 2xz = x \Leftrightarrow \left[ \begin{array}{l}x = 0 \\ x + 2z = 1 \\ \end{array} \right.$
$x = y = 0:$
$\left( 3 \right) \Leftrightarrow z^2 = z \Leftrightarrow \left[ \begin{array}{l}z = 0 \\ z = 1 \\ \end{array} \right.$
$x + 2z = 1 \Leftrightarrow x = 1 - 2z $
$\left( 3 \right) \Leftrightarrow z^2 + 2\left( {1 - 2z} \right)^2 = z \Leftrightarrow 9z^2 - 9z + 2 = 0 \Leftrightarrow \left[ \begin{array}{l}z = \dfrac{2}{3} \\ z = \dfrac{1}{3} \\ \end{array} \right. $
$\Rightarrow \left[ \begin{array}{l}x = y = \dfrac{{ - 1}}{3} \\ x = y = \dfrac{1}{3} \\ \end{array} \right.$
$\bullet x + y = 2z + 1 \Leftrightarrow 2z = x + y - 1 $
$\left( 1 \right) + \left( 2 \right) x^2 + y^2 + 2z\left( {x + y} \right) = x + y \Leftrightarrow 2\left( {x + y} \right)^2 - 2xy - 2\left( {x + y} \right) = 0 $
$\Leftrightarrow \left( {x + y} \right)^2 = xy + x + y\left( 4 \right) $
$\left( 3 \right) \Leftrightarrow \left( {\dfrac{{x + y - 1}}{2}} \right)^2 + 2xy = \dfrac{{x + y - 1}}{2} \Leftrightarrow x^2 + y^2 + 1 + 2xy - 2x - 2y + 8xy = 2\left( {x + y - 1} \right)$
$\Leftrightarrow \left( {x + y} \right)^2 - 4\left( {x + y} \right) + 8xy + 3 = 0\left( 5 \right) $
$\left( 4 \right),\left( 5 \right) \Rightarrow 9xy - 3\left( {x + y} \right) + 3 = 0 \Leftrightarrow 3xy + 1 = x + y $
$\Leftrightarrow xy = \dfrac{{x + y - 1}}{3} \Rightarrow \left( {x + y} \right)^2 = \dfrac{{4\left( {x + y} \right) - 1}}{3} $
$t = x + y \Rightarrow 3t^2 - 4t + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 1 \\ t = \dfrac{1}{3} \\ \end{array} \right. \Rightarrow \left[ \begin{array}{l}z = 0 \\ z = \dfrac{{ - 1}}{3} \\ \end{array} \right.$
$\left( 3 \right) \Rightarrow \left[ \begin{array}{l}xy = 0 \\ xy = \dfrac{{ - 2}}{9} \\ \end{array} \right. $
$\left\{ \begin{array}{l}xy = 0 \\ x + y = 1 \\ \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l}x = 0 \\ y = 1 \\ \end{array} \right. \\ \left\{ \begin{array}{l}x = 1 \\ y = 0 \\\end{array} \right. \\ \end{array} \right. $
$\left\{ \begin{array}{l}xy = \dfrac{{ - 2}}{9} \\ x + y = \dfrac{1}{3} \\ \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = \dfrac{2}{3} \\ y = \dfrac{{ - 1}}{3} \\ \end{array} \right. \\\left\{ \begin{array}{l}x = \dfrac{{ - 1}}{3} \\ y = \dfrac{2}{3} \\ \end{array} \right. \\ \end{array} \right. $
$\Rightarrow \left( {x,y,z} \right) = \left( {0,0,0} \right);\left( {0,0,1} \right);\left( {\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{3},\dfrac{2}{3}} \right);\left( {\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}} \right);\left( {0,1,0} \right);\left( {1,0,0} \right);\left( {\dfrac{2}{3},\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{3}} \right);\left( {\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 1}}{3}} \right) $

Bài viết đã được chỉnh sửa nội dung bởi dark templar: 18-12-2010 - 21:57