Chủ đề này có 2 trả lời

[bong2000]

Bài 1: Cho $a+b+c=0;$

$x+y+z=0;$

$bcx+cay+abz=0;$

 Tính $a^{2}x+b^{2}y+c^{2}z$

Bài 2: Cho 

$\left ( x-y \right )^{2}+\left ( y-z \right )^{2}+\left ( z-x \right )^{2}=\left ( z+x-2y \right )^{2}+\left ( z+y-2x \right )^{2}+\left ( x+y-2x \right )^{2}$

 Chứng minh rằng : $x=y=z$

Bài viết đã được chỉnh sửa nội dung bởi Oral1020: 06-06-2013 - 12:23

[sieumau88]

Bài 1: Cho_ $a+b+c=0 \Leftrightarrow a = - b - c$

$x+y+z=0 \Leftrightarrow -y = x + z$ _và_ $-z = x + y$

$bcx+cay+abz=0 \Leftrightarrow bcx = - cay - abz$

 Tính_ $a^{2}x+b^{2}y+c^{2}z$

 

Ta có

 

$M = a^2x + b^2y + c^2z = (-b-c)^2x + b^2y + c^2z$

$M = b^2x + c^2x + 2bcx + b^2y + c^2z$

$M = b^2 (x + y) + c^2 (x+z) + bcx + bcx$

$M = -b^2z - c^2y - cay - abz + bcx$

$M = -bz (b+a) - cy (c+a) + bcx$

$M = bz c + cy b + bcx$

$M = bc .(z+y+x)$

$M = bc . 0$

$M =  0$

[sieumau88]

Bài 2: Cho 
$\left ( x-y \right )^{2}+\left ( y-z \right )^{2}+\left ( z-x \right )^{2}=\left ( z+x-2y \right )^{2}+\left ( z+y-2x \right )^{2}+\left ( x+y-2z \right )^{2}$ (**)
 
Chứng minh rằng : $x=y=z$

 
Ta có
$VP = \left [ (z-y)+(x-y) \right ]^{2} + \left [ (z-x)+(y-x) \right ]^{2} + \left [ (x-z)+(y-z) \right ]^{2}$
 
$VP = 2\left (x-y \right )^{2} + 2\left (y-z \right )^{2} + 2\left (z-x \right )^{2} + 2 \left (z-y \right)\left (x-y \right) + 2 \left (z-x \right)\left (y-x \right) + 2 \left (x-z \right)\left (y-z \right)$
 
Đặt $\left\{\begin{matrix}
a = x-y \\
b = y-z\\
c = z-x
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
a + b = -c\\
b + c = -a\\
c + a = -b
\end{matrix}\right.$
 
$VP = 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc$
$VP = 2a^2 + 2b^2 + 2c^2 - a(b+c) - b(a+c) - c(a+b)$
$VP = 2a^2 + 2b^2 + 2c^2 + a^2 + b^2 + c^2$
$VP = 3(x-y)^2 + 3(y-z)^2 + 3(z-x)^2$

khi đó (**) $\Leftrightarrow \left ( x-y \right )^{2}+\left ( y-z \right )^{2}+\left ( z-x \right )^{2}=3\left ( x-y \right )^{2}+3\left ( y-z \right )^{2}+3\left ( z-x \right )^{2}$
$\Leftrightarrow 2\left ( x-y \right )^{2}+2\left ( y-z \right )^{2}+2\left ( z-x \right )^{2} = 0$
$\Leftrightarrow \left\{\begin{matrix}
x-y = 0\\
y-z = 0\\
z-x = 0
\end{matrix}\right.$
Vậy $x = y = z$