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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x+y=0\) 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 !!

pt 1) x=y=z  Cosi 3 số 

Xét với \(0< x,y,z< 1\) thì \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}>1\) (vô lí)

Xét \(x,y,z\ge1\) , đặt \(\hept{\begin{cases}x=a^3\\y=b^3\\z=c^3\end{cases}}\) (\(a,b,c\ge1\))

Ta có \(1=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{1}{a^3+1}+\frac{1}{b^3+1}+\frac{1}{c^3+1}\ge\frac{3}{abc+1}\) (cái này chắc you cm đc)

\(\Rightarrow abc\ge2\Rightarrow a^3.b^3.c^3\ge8\) hay \(xyz\ge8\) (1)

Áp dụng BĐT AM-GM : \(1=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{9}{x+y+z+3}\Rightarrow x+y+z\ge6\) (2)

Áp dụng BĐT Cauchy : \(1=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{3}{\sqrt[3]{\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\) 

\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge27\) (3)

Nhân (1), (2), (3) theo vế : \(xyz\left(x+y+z\right)\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge1296\)

Đẳng thức xảy ra khi xảy ra đồng thời (1), (2), (3) , tức là x = y = z = 2

Vậy tập nghiệm của hệ : \(\left(x,y,z\right)=\left(2;2;2\right)\)

you chứng minh \(xyz\ge8\) thử coi được không?

\(P=\frac{1}{xy-xyz-z}+\frac{1}{yz-xyz-x}+\frac{1}{xz-xzy-y}\)  .Do xyz=-z =>-xyz=1 và x+y+z=0 . Thế vào P ta được \(P=\frac{1}{xy+1+x+y}+\frac{1}{yz+1+y+z}+\frac{1}{xz+1+x+z}\)\(P=\frac{1}{\left(x+1\right)\left(y+1\right)}+\frac{1}{\left(y+1\right)\left(z+1\right)}+\frac{1}{\left(x+1\right)\left(z+1\right)}\) =\(\frac{z+1+x+1+y+1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\) 

\(P=\frac{3}{xyz+z+xz+yz+xy+1+x+y}\) =\(\frac{3}{xy+yz+xz}\) (Do x+y+z=0; xyz=-1)

x+y+z=0 => (x+y+z)²=0 => x²+y²+z² +2(xy+yz+xz)=0 => 2(xy+yz+xz)=-6 => xy+yz+xz=-3 Thế vào P ta được :

\(P=\frac{3}{-3}=-1\) . Chúc bạn học tốt

Hình như bạn ghi thiếu đề r . Còn xyz=-1 nữa 

Đặt \(\sqrt{x}=x;\sqrt{y}=y;\sqrt{z}=z\) cho dễ nhìn.

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\x^2+y^2+z^2=2\end{cases}}\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\)

\(\Leftrightarrow xy+yz+zx=1\)

Ta có:

\(x\left(1+y^2\right)\left(1+z^2\right)+y\left(1+z^2\right)\left(1+x^2\right)+z\left(1+x^2\right)\left(1+y^2\right)\)

\(=x^2y^2z+y^2z^2x+z^2x^2y+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+x+y+z\)

\(=xyz\left(xy+yz+zx\right)+x^2\left(2-x\right)+y^2\left(2-y\right)+z^2\left(2-z\right)+2\)

\(=-2xyz+2\left(x^2+y^2+z^2\right)-\left(x^3+y^3+z^3-3xyz\right)+2\)

\(=-2xyz+6-\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=-2xyz+6-2=-2xyz+4\)

Ta lại có:

\(\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\)

\(=x^2y^2z^2+\left(xy+yz+zx\right)^2-2xyz\left(xy+yz+zx\right)+3\)

\(=x^2y^2z^2-2xyz+4=\left(xyz-2\right)^2\)

\(\Rightarrow A=\sqrt{\left(xyz-2\right)^2}.\frac{4-2xyz}{\left(xyz-2\right)^2}\)

Tới đây bí :((

thanks nha, z là ok rồi