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(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0
(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0
(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0
(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0
Suy ra x+y+z =0
x+y = -z
y+z = -x
x+z = -y
B = -16 + (-3) +2038 = 2019
Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)
+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)
\(=-16-3+2038=2019\)
+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)
\(=32+6-4076=-4038\)
\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z-xy^2+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left[xy-zx-zy+z^2\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Vậy A = 1
\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}=\frac{\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)}=\frac{1}{x-y}-\frac{1}{x-z}\)
\(\frac{z-x}{\left(y-z\right)\left(y-x\right)}=\frac{\left(y-x\right)-\left(y-z\right)}{\left(y-z\right)\left(y-x\right)}=\frac{1}{y-z}-\frac{1}{y-x}\)
\(\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{\left(z-y\right)-\left(z-x\right)}{\left(z-x\right)\left(z-y\right)}=\frac{1}{z-x}-\frac{1}{z-y}\)
Suy ra: \(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}\)
\(=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
rồi bí mẹ chỗ này
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
\(A=\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^3\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{y^3\left(z-x\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{z^3\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^3\left(y-z\right)-y^3\left(x-y+y-z\right)+z^3\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^3\left(y-z\right)-y^3\left(x-y\right)-y^3\left(y-z\right)+z^3\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x^3-y^3\right)-\left(x-y\right)\left(y^3-z^3\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(y-z\right)\left(y^2+z^2+zy\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x^2+xy+y^2-y^2-z^2-zy\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left[\left(x^2-z^2\right)+\left(xy-zy\right)\right]}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x+z\right)+y\left(x-z\right)\right]}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+z+y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=x+y+z=2020\)