phân tích gt sau đó suy ra x+y+x=0 

từ đây tính đc x+y=? y+z=? x+z=? 

ta được kết quả là'; -2006

Xét \(x^3+y^3+z^3=3xyz\)

\(x^3+y^3+z^3-3xyz=0\)

\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)

\(\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+2xy+y^2-xy-yz+z^2\right)-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

TH1:\(x+y+z=0\) 

\(\Rightarrow x+y=-z;y+z=-x;z+x=-y\left(1\right)\)

Thay (1) vô pt cần tính:

\(\frac{2016xyz}{-z.-x.-y}=\frac{2016xyz}{-\left(xyz\right)}=-2016\)

TH2:\(x^2+y^2+z^2-xy-yz-xz=0\)

Nhân 2 vế với 2

\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

Do VT dương

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}}\Rightarrow x=y=z\)

Thay y,z ở pt cần tính là x

\(\Rightarrow\frac{2016x.x.x}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=\frac{2016}{8}=252\)

Vậy pt có thể = -2016 khi x + y + z = 0

       pt có thể = 252 khi \(x^2+y^2+z^2-xy-xz-yz=0\)

\(x^3+y^3+z^3=3xyz\)

\(x^3+y^3+z^3-3xyz=0\)

\(x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0\)

\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-xz-yz=0\end{cases}}\)

TH1 : \(x+y+z=0\)

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

Thay vào biểu thức ta có :

\(\frac{2016xyz}{\left(-z\right)\left(-x\right)\left(-y\right)}=\frac{2016xyz}{-xyz}=-2016\)

TH2 : \(x^2+y^2+z^2-xy-xz-yz=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+\left(y^2-2yz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}\Rightarrow}x=y=z}\)

Từ đây ta biến đổi biểu thức :

\(\frac{2016x^3}{2x\cdot2x\cdot2x}=\frac{2016x^3}{8x^3}=252\)

Vậy........

=-2016 đúng ko?

Đề chưa chuẩn: tuy nhiên đánh vào -2016 => đáp án đúng:

Vì bản chất như sau:

thỏa ĐK ban đầu x^3+y^3+z^3=3xzy

Từ HĐT=>

\(\orbr{\begin{cases}x+y+z=0\left(1\right)\\x^2+y^2+z^2-xy-yz-xz=0\left(2\right)\end{cases}}\)

=>(1)&(2) đều có cặp nghiệm x=y=z=0 khi đó P không xác định

do vậy đề thiếu điều kiện x,y,z không đồng thời =0:(*)

Nếu thêm đk (*) giải tiếp

(2) vô nghiệm 

do vậy khi đó chỉ có nghiệm duy nhất của (1) 

x+y=-z

x+z=-y

z+y=-x

Thay vào biểu thwucs  P=-2016

Đây cũng là lý do mình không thích thi VIOLIMPIC 

nhiều đáp án sai--> đề sai--> gây ức chế--> không thi

\(\left(x^3+1\right)\left(y^3+1\right)\left(z^3+1\right)=\dfrac{81}{64}x^3y^3z^3\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^2y^2z^2\)

\(\Leftrightarrow3xyz\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^3y^3z^3\)

 \(\Rightarrow\left[{}\begin{matrix}xyz=0\\\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\end{matrix}\right.\)

Nếu \(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\) 

Ta có:

\(x^2-x+1=\dfrac{3}{4}x^2+\left(\dfrac{x}{2}-1\right)^2\ge\dfrac{3}{4}x^2\)

Tương tự: \(y^2-y+1\ge\dfrac{3}{4}y^2\) ; \(z^2-z+1\ge\dfrac{3}{4}z^2\)

Do các vế của các BĐT trên đều không âm, nhân vế với vế ta được:

\(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge\dfrac{27}{64}x^2y^2z^2\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\) 

Thế vào  điều kiện \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=3xyz\) ko thỏa mãn (loại)

Vậy \(xyz=0\)

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)

GIÁO VIÊN SAO TOÀN SAI HẰNG ĐẲNG THỨC THẾ????

\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)

@@ ko ra nữa

Ta có:

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

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Rightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Rightarrow\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)=0\)

\(\Rightarrow\left(x+y+z\right)^3-\left(x+y+z\right)\left(3xy+3zx+3yz\right)=0\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xy-3xz-3yz\right)=0\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Rightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)

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

\(\Rightarrow x=y=z\)

Xét trường hợp x = y = z, ta có:

\(P=\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(P=\dfrac{x^3}{2x.2x.2x}\)

\(P=\dfrac{x^3}{8x^3}\)

\(P=\dfrac{1}{8}\)

Xét trường hợp x + y + z = 0, ta có:

\(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{-\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(\Rightarrow P=-1\)

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

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy.\left(x+y+z\right)\)

\(=\left(x+y\right)^2-\left(x+y\right).z+z^2-3xy\left(\text{vì }x+y+z=1\right)\)

\(=x^2+2xy+y^2-xz-yz+z^3-3xy\)

\(=x^2+y^2+z^2-xy-yz-xz\)

\(=\frac{1}{2}.\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)\)

\(=\frac{1}{2}.\left[\left(x^2-2xy-y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\right]\)

\(=\frac{1}{2}.\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)=VP

=>dpcm

Ta có : \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

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

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

\(=x^2+y^2+z^2-xy-yz-xz=\frac{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)}{2}=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)