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........
Ta có: \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\) hoặc \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)
-Nếu \(x+y+z=0\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)
Suy ra \(\frac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2016xyz}{-xyz}=-2016\)
-Nếu \(x=y=z\)
Suy ra: \(\frac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=252\)
Vậy ......
