+> Lấy (x + y + z)^2 = x^2+y^2+z^2+2xy+2yz+2xz = 1+2xy+2yz+2xz

Mà (x + y + z)^2 = 1

=> 2xy+2yz+2xz = 0

=> xy+yz+xz = 0

=> (xy+yz+xz)(x + y + z) = 0

+> Lấy (x + y + z)^3 = x^3 + y^3 + z^3 + 6xyz + 3xy^2 + 3x^2y + 3x^2z + 3xz^2 + 3yz^2 + 3y^2z = 1 +  6xyz + 3xy^2 + 3x^2y + 3x^2z + 3xz^2 + 3yz^2 + 3y^2z 

Mà (x + y + z)^3 = 1

=>  6xyz + 3xy^2 + 3x^2y + 3x^2z + 3xz^2 + 3yz^2 + 3y^2z = 0

=> 6xyz + 3(xy^2 + x^2y + x^2z + xz^2 + yz^2 + y^2z) = 0

=> 6xyz + 3[xy(x+y) + xz(x+z) + yz(y+z)] = 0

=> 6xyz + 3[xy(1-z) + xz(1-y) + yz(1-x)] = 0

=> 6xyz + 3(xy - xyz + xz - xyz + yz - xyz) = 0

Mà xy+yz+xz = 0

=> 6xyz - 9xyz = 0

=> xyz = 0

Mà (xy+yz+xz)(x + y + z) = 0

=> (xy+yz+xz)(x + y + z) = xyz

=> (xy+yz+xz)(x+y+z) - xyz = 0

Phân tích đa thức trên thành nhân tử, ta có (x+y)(y+z)(x+z) = 0

=> x+y = 0 ; y+z = 0 ; x+z = 0

Có x^2017 + y^2017 + z^2017

= (x+y)(x^2017 -x^2016y+...+y^2017) + z^2017         (1)

= z^ 2017
Có x+y = 0 => x = -y

=> (x + y + z )^2017 = z^2017                                  (2)

Từ (1) và (2) = > x^2017 + y^2017 + z^2017 = (x + y + z )^2017 = 1

x = 1, y= 0

hoặc x = 0, y = 1

mình cũng chẳng biết đúng hay sai

\(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)

<=> x+y = 0 hoặc x+z=0 hoặc z+y=0

<=> x = -y hoặc x = -z hoặc z = -y

\(\Rightarrow P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)

 với mọi x, y, z ta có: 

(x-y)^2 +(y-z)^2+ (z-x)^2>=0 

<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0 

<=>x^2 + y^2 +z^2 - xy -yz -zx >=0 

<=>(x+y+z)^2 >= 3(x+y+z) 

<=>[(x+y+z)^2]/3 >= xy+yz+ zx 

=>xy +yz + zx <=3 

dấu = xảy ra khi x=y=z =1

hình như bài của mik làm có j đó sai sai

với mọi x, y, z ta có: 

(x-y)^2 +(y-z)^2+ (z-x)^2>=0 

<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0 

<=>x^2 + y^2 +z^2 - xy -yz -zx >=0 

<=>(x+y+z)^2 >= 3(x+y+z) 

<=>[(x+y+z)^2]/3 >= xy+yz+ zx 

=>xy +yz + zx <=3 

dấu = xảy ra khi x=y=z =1

với mọi x, y, z ta có:

(x-y)^2 +(y-z)^2+ (z-x)^2>=0

<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0

<=>x^2 + y^2 +z^2 - xy -yz -zx >=0

<=>(x+y+z)^2 >= 3(x+y+z)

<=>[(x+y+z)^2]/3 >= xy+yz+ zx

=>xy +yz + zx <=3

dấu = xảy ra khi x=y=z =1 

\(\left\{{}\begin{matrix}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)

\(\Leftrightarrow x+y=0\) hoặc \(x+z=0\) hoặc \(z+y=0\)

\(\Leftrightarrow x=-y\) hoặc \(x=-z\) hoặc z=-y

\(\Rightarrow P\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)

Chúc bạn học tốt !!

!x-2007!+!x-2010!>=3 đẳng thức khi 2007<=x<=2008 

!x-2007!+!x-2008!+!x-2010!>=3 đẳng thức khi !x-2008!=0 

=> nghiệm duy nhất x=2008 và y=2009