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Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)
\(A=\frac{\left(xy+2016z\right)\left(yz+2016x\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)
Thay \(x+y+z=2016\)
\(A=\frac{\left[xy+\left(x+y+z\right)z\right]\left[yz+\left(x+y+z\right)x\right]\left[zx+\left(x+y+z\right)y\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)
\(A=\frac{\left[xy+xz+yz+z^2\right]\left[yz+xy+xz+x^2\right]\left[zx+xy+yz+y^2\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left[x\left(y+z\right)+z\left(y+z\right)\right]\left[y\left(z+x\right)+x\left(z+x\right)\right]\left[x\left(z+y\right)+y\left(z+y\right)\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left[\left(y+z\right)\left(x+z\right)\right]\left[\left(x+z\right)\left(x+y\right)\right]\left[\left(z+y\right)\left(x+y\right)\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left(x+z\right)\left(x+z\right)\left(y+z\right)\left(y+z\right)\left(x+y\right)\left(x+y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left(x+z\right)^2\left(y+z\right)^2\left(x+y\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=1\)
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
b: 5x^2+5y^2+8xy-2x+2y+2=0
=>4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0
=>(x-1)^2+(y+1)^2+(2x+2y)^2=0
=>x=1 và y=-1
M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1